UC-NRLF 


SB    27fl 


1.  7L 


IN  MEMORIAM 
FLOR1AN  CAJORI 


TEIGONOMETEY 


FOR 


SCHOOLS  AND  COLLEGES 


FREDERICK   AKDEREGG,   A.M. 

PROFESSOR  OP  MATHEMATICS 


AND 

EDWARD   DRAKE   ROE,  JR.,  A.M. 

ASSOCIATE  PROFESSOR  OF  MATHEMATICS 

IN  OBERLIN  COLLEGE 


BOSTON,  U.S.A.,  AND  LONDON 

GINN    &    COMPANY,    PUBLISHERS 

1896 


COPYRIGHT,  1896,  BY 
F.  ANDEREGG  AND  E.  D.  HOE,  JR. 


ALL  BIGHTS  RESERVED 


,  CAJORI 


PKEFACE. 


THIS  book  is  a  revision  of  lectures  on  trigonometry  which, 
we  have  given  to  our  students  for  several  years,  and  have 
furnished  them  during  the  last  two  years  in  manifolded  form. 
The  results  have  been  so  satisfactory,  that  we  have  been  led 
to  hope  that  the  work  would  prove  useful  in  a  larger  field. 
Our  aim  has  been  to  adapt  the  work  to  the  needs  of  students 
and  teachers,  with  reference  to  other  mathematical  subjects 
both  elementary  and  advanced. 

Since  in  our  own  and  a  growing  number  of  other  institu- 
tions, an  explicit  course  on  a  book  of  tables  precedes  the 
course  in  trigonometry,  we  have  omitted  that  work,  and  do  not 
publish  tables.  We  have  used  some  fundamental  principles 
of  algebra  and  geometry,  with  which  students  of  trigonometry 
can  be  assumed  to  be  familiar,  without  giving  references. 

We  have  not  given  numerical  solutions  of  examples  under 
the  different  cases  of  triangles,  because  the  competent  teacher 
does  not  need  them,  and  we  wish  to  leave  him  free  to  use 
such  models  as  he  prefers. 

We  have  endeavored  to  lay  down  general  conventions  and 
definitions,  to  emphasize  consistency  in  the  use  of  the  con- 
ventions, to  give  general  demonstrations  in  which  we  have 
carefully  aimed  at  logical  soundness,  directness,  and  sim- 
plicity, and  to  exhibit  the  unity  of  the  subject  as  made  up  of 


911316 


iv  PREFACE. 

its  related  parts.  We  believe  that  in  this  way  the  work  is 
simplified,  the  student  gets  the  ground  plan  of  the  higher 
analysis,  and  is  saved  from  much  subsequent  intellectual 
lameness,  which  results  from  an  excessive  absorption  of 
the  attention  on  special  cases.  At  the  same  time,  when  it 
seemed  pedagogically  helpful,  we  have  led  up  to-  the  general 
demonstration  by  the  consideration  of  special  cases. 

We  have  given  numerous  examples  at  appropriate  points, 
in  order  to  admit  of  such  selection  as  the  teacher  may  find 
necessary  for  different  classes,  and  to  secure  a  thorough 
application  of  the  theory. 

The  reader  will  find  a  number  of  original  features  in  the 
book. 

Our  views  on  trigonometry  and  related  subjects  have  been 
influenced  by  many  American,  English,  French,  and  German 
treatises,  and  especially  by  our  respected  teachers,  Professors 
J.  M.  Peirce,  W.  E.  Byerly,  and  B.  0.  Peirce,  of  Harvard 
University. 

The  figures  which  appear  in  this  book  were  drawn  by 
Mr.  D.  C.  Churchill  of  this  college.  We  have  taken  some 
examples  from  other  text-books. 

F.  A.  AND  E.  D.  E.,  JR. 


CONTENTS. 

PLANE   TRIGONOMETRY. 
CHAPTER   I. 

PAGE 

A.  LINEAR  AND  ANGULAR  MAGNITUDE     ......  1 

DEFINITION  OF  TRIGONOMETRY        ......  2 

B.  MEASURE-UNITS .3 

a.  Linear  Units  and  their  Transformations         ...  3 

6.   Angular  Units  and  their  Transformations           ...  3 

1.  Sexagesimal ;  2.  Centesimal ;  3.  Circular    .        .  3 

C.  EXAMPLES.    1 5 

CHAPTER   II. 

A.  CONVENTIONS  ..........  6 

a.  Conventions  about  Lines 6 

6.   Conventions  about  Angles 7 

B.  TRIANGLE  OF  REFERENCE    ........  7 

C.  TRIGONOMETRIC  FUNCTIONS     .......  8 

_D.    MUTUAL  RELATIONS  OF  FUNCTIONS     .         .         .         .         .         .  11 

E.    EXAMPLES.    II 13 

CHAPTER   III. 

A.  VARIATION  OF  TRIGONOMETRIC  FUNCTIONS  .         .         .         .16 

B.  LINE  REPRESENTATIONS  OF  THE  TRIGONOMETRIC  FUNCTIONS  20 

C.  INVERSE  TRIGONOMETRIC  FUNCTIONS 22 

D.  EXAMPLES.    III. 


CONTENTS. 


CHAPTER   IV. 


A.  TRIGONOMETRIC     FUNCTIONS     OF     INTEGRAL     MULTIPLES     OP 

-,  ±  0,  i.e.,  of  m  -  ±  <f>  .        .         .       -.  .         .         .24 

a.  When  m  is  even,  and  equal  to  2  n  .         .         .  .         .   -      32 

6.   When  m  is  odd,  and  equal  to  (2  n  +  1)       .  .         .         .33 

B.  PERIODICITY  OP  THE  TRIGONOMETRIC  FUNCTIONS  .  .         .36 

C.  THE  FUNCTIONS  OP  SOME  SPECIAL  ANGLES         .  .         .         .37 

D.  EXAMPLES.    IV.  38 


CHAPTER   V. 

A.  PROJECTION          .         ....        t        ....  42 

a.  When  the  Line  and  Axis  are  Co-planar          ...         .  42 
6.   When  the  Line  and  Axis  are  not  Co-planar         .         .         .43 

B.  FUNCTIONS  OP  SUMS  AND  DIFFERENCES  OF  ANGLES       .         .  44 

C.  FUNCTIONS  OF  MULTIPLE  ANGLES        ......  46 

D.  FUNCTIONS  OF  SUB-MULTIPLE  ANGLES    .         ...         .         .  46 

E.  FORMULAS  FOR  THE  SUMS  AND  DIFFERENCES  OP  THE  SINES  OP 

TWO  ANGLES,  ALSO  OF  THE  COSINES  OF  TWO  ANGLES      .  47 

F.  EXAMPLES.    V .        .         .         .         .         .     '    ,         .         .         .  48 

G.  EXAMPLES.    VI.  49 


CHAPTER  VI. 

A.  RIGHT  TRIANGLES 52 

a.  Cases  ..........  52 

b.  Solutions 52 

B.  EXAMPLES.    VII.  , 53 

C.  PRACTICAL  APPLICATIONS  TO  PROBLEMS  ON  HEIGHTS  AND  DIS- 

TANCES  ..........  54 

D.  EXAMPLES.    VIII 64 


CONTENTS.  vjj 


CHAPTER   VII. 

PAGE 

A.  SOME  GENERAL  FORMULAS  FOR  OBLIQUE  TRIANGLES     .         .  56 

a.  Theorem  of  Sines 56 

b.  Theorem  of  Tangents     .......  57 

c.  Theorem  of  Cosines     ........  57 

d.  Formulas  for  one  Side  of  a  Triangle  in  Terms  of  the  Adja- 

cent Angles  and  the  other  two  Sides           ...  58 
A     R     c* 

e.  Formulas  for  the  Functions  of  - ,  - ,  -      .         .         .         .58 

/.  Formulas  for  the  Area  of  a  Triangle,  and  the  Radii  of  the 

Inscribed  and  Circumscribed  Circles  ....  60 

B.  OBLIQUE  TRIANGLES 62 

a.  Cases 62 

6.    Solutions     .         .         . 63 

C.  EXAMPLES.    IX.  64 


SPHERICAL   TRIGONOMETRY. 

CHAPTER   I. 

A.  INTRODUCTION 67 

a.  Geometrical  Principles   .......          67 

6.   Sufficiency  of  the  Convex  Spherical  Triangle       .         .         .68 

B.  GENERAL  THEORY  OF  SPHERICAL  TRIGONOMETRICAL  RELATIONS     69 

C.  SYSTEMS  OF  EQUATIONS  DEMANDED  BY  THE  PRECEDING  THEORY     70 

a.  Equations  in  which  the  Parts  enter  Four  at  a  Time  .  70 
6.  Equations  in  which  the  Parts  enter  Five  at  a  Time  .  .  75 
c.  Equations  in  which  the  Parts  enter  Six  at  a  Time  .  76 

D.  ON  THE  USE  OF  THE  POLAR  TRIANGLE  IN  ESTABLISHING  RELA- 

TIONS           .         .         .         .76 


Viii  CONTENTS. 


CHAPTER   II. 

PAGE 

A.  FORMULAS        .         ...         .         .         .         .         .  .          78 

a.  Spherical  Right  Triangles  .        ...        .        .  .78 

6.   Quadrantal  Triangles .          78 

c.   Deductions          .        .        '.      •  .        .        .        .        .  .79 

B.  SOLUTIONS        ..........  79 

a.  Right  Triangle    .        .     -..'-.•      .        .        .        .  .79 

6.   Quadrantal  Triangle       .        .     ^.        ...  .          79 

C.  SOME  ASTRONOMICAL  CONCEPTIONS      ...       .         .       ...  .80 

a.  Celestial  Sphere      .        .        ...        .        .  .          80 

b.  Spherical  Coordinates          ...       .        .        .        .        .  .82 

D.  EXAMPLES.    X.  84 


CHAPTER   III. 

A.  FORMULAS  FOR  SPHERICAL  OBLIQUE  TRIANGLES          .         .         .87 

a.  Formulas  for  Functions  of  |  A,  $  B,  £  C        .        -.         .          87 

6.  Gauss's  Equations       . 88 

c.  Napier's  Analogies       .  .        .-        .        ...        .        .          90 

d.  1'Huilier's  Formula     .         .  .         .         .         .91 

e.  Formula  for  the  Polar  Radius  of  the  Circumscribed  Circle 

of  a  Spherical  Triangle 92 

/.  Formula  for  the  Polar  Radius  of  the  Inscribed  Circle  of  a 
Spherical  Triangle     .        . 92 

B.  PLANE  TRIGONOMETRY  A  SPECIAL  CASE  OF  SPHERICAL  TRIGO- 

NOMETRY         ....,»..  .          93 

C.  SPHERICAL  TRIANGLES         .         .         .         .         .         .         .         .94 

a.  Cases      .,...' 94 

6.   Solutions     .         .        .        .        .         .         .         .         .         .95 

D.  EXAMPLES.    XI.  106 


PLANE    TEIGONOMETRY. 


CHAPTER   I. 

A.    Linear  and  Angular  Magnitude.    Definition  of 
Trigonometry. 

§  1.  Magnitude  is  duration  in  time,  or  extent  of  space,  or 
duration  in  time  and  extent  of  space  of  that  which  manifests 
itself  in  time  and  space. 

§  2.  A  line  is  the  path  of  a  moving  point.  It  has  one 
dimension  of  space,  length.  The  initial  position  of  the  gen- 
erating point  is  the  origin  of  the  line.  The  final  position  is 
its  term.  The  line  is  taken  from  the  origin  to  the  term,  and 
is  always  determinate,  unless  the  origin  and  term  coincide, 
when  it  may  be  indeterminate.  Hence  every  line  joining 
two  points  represents  at  least  two  lines,  opposites  in  direc- 
tion, and  negatives  of  each  other.  That  is  : 

A  B 


It  follows  from  this  that  if  AI,  A2, An,  be  n  points  arranged 

in  any  order  on  a  line, 

AiA2+A2A3+.A8A4  + A^An+AA^I, 

where  k  is  either  zero,  or  any  positive  or  negative  integer, 
and  I  is  the  simplest  length  of  the  line.  Linear  magnitude 
is  the  extent  of  a  line,  and  may  be  unlimited. 


'2' 


PLANE   TRIGONOMETRY. 


aiigle  is  the  amount  of  turning  in  a  plane  of  a 
straight  line  about  a  fixed  point  through  which  it  passes. 
The  initial  position  of  the  generating  line  is  called  an  axis 
of  reference,  and  that  part  of  it  which  forms  the  initial  side 
of  the  angle  is  called  the  initial  line  of  the  angle.  The 
whole  of  the  generating  line  is  called  the  rotating  line,  and 
that  part  of  it  which  forms  the  terminal  side  of  the  angle  is 
called  the  terminal  line  of  the  angle.  The  angle  is  taken 
from  the  initial  to  the  terminal 
line.  Hence  every  geometric 
angle  must  represent  at  least  two 
angles,  opposites  in  turning,  and 
negatives  of  each  other.  That  is  : 
AOB+BOA^O,  or 

BOA  =  -AOB 
AOB  =  —  BOA. 
It  follows  from  this  that  if  ax,  a2, an,  be  any  n  lines  in  a 

plane,  and    '  represent  the  angle  from  ak  to  a\ .  then 
ak 

a  ~^~  a  ~^~  a  ~^~ a        +  ^  =  "360°, 

where  n  may  equal  zero  or  any  integer,  positive  or  negative. 
Angular  magnitude  is  the  amount  of  turning  of  the  generating 
line,  and  may  be  unlimited. 

§  4.  It  is  to  be  noticed  that  the  preceding  definitions  of  a 
line,  and  an  angle,  are  extensions  of  the  elementary  concep- 
tions ;  for  they  require  as  essential  to  the  line,  and  the  angle, 
two  elements,  magnitude  and  direction.  The  elementary 
conceptions  require  only  magnitude.  From  these  definitions 
corresponding  extensions  of  all  the  elementary  conceptions 
into  which  lines,  and  angles  enter,  necessarily  follow. 

§  5.  The  subject  matter  of  Trigonometry  is  lines,  and  angles. 
Trigonometry  is  the  investigation  of  the  relations  of  the  sides 
and  angles  of  a  triangle. 


LINEAR  AND  ANGULAR  MAGNITUDE. 


B.  Measure-Units. 

§  6.  The  measure  of  a  magnitude  is  expressed  by  the 
number  of  measure-units  the  magnitude  contains.  Thus,  the 
measure  of  a  line  is  3  feet,  4  rods,  2  meters,  etc.  Both  lines 
and  angles  are  measured  by  different  measure-units. 

(a)   Linear  Units  and  their  Transformation. 

§  7.   The  principal  measure-units  for  lines  are  the  English 
and  the  French.     The  equation  for  transformation  is  : 
1  meter  ^39.37  inches,  approximately. 

(6)   Angular  Units  and  their  Transformations. 

§  8.  The  principal  measure-units  for  angles  are  the  sexa- 
gesimal, the  centesimal,  and  the  circular. 

1.  In  the  sexagesimal  system  : 

1  right  angle  =  90  degrees,  written  90°. 
1°  =  60  minutes,       "       60'. 

1'  =60  seconds,       "       60". 

2.  In  the  centesimal  system  : 

1  right  angle  =  100  grades,  written  100g. 
1«  =100  minutes,     «       100V. 

T  =100  seconds,      «       100". 

3.  The  radian  is  that  central 
angle   in   any   circle   whose    in- 
tercepted arc  is.  equal  in  length 
to  the  radius.    Since  in  the  same 
circle,  or  in  equal  circles,  central 
angles  are  proportional  to  their 
intercepted  arcs,  and  if  8  repre- 
sents the  number  of  degrees  in 
the  angle,  r  the  radius,  and  s  the 

arc,  then  :  FIG.  i. 


PLANE  TRIGONOMETRY. 

«_.  _!_  a} 

360      2  Trr ' 

=  ilo-8-  V 

\     s 

This  shows  that  f  -  J  varies  as  8,  that  is,  varies  as  the  angle. 
Let  <f>  represent  the  angle  in  any  system  of  measurement, 

then  d>  =  n  (  -  ) ,  where  n  is  constant.     When  s  =  r,  d>  is  con- 

W 
stant ;  also  when  s  =  r,  <£  is  a  radian,  which  we  will  take  as 

the  angular  measure-unit  in  the  circular  system,  and  represent 
by  r  written  like  an  exponent ;  and  we  have,  to  determine  n, 

1  =  n  (  -  ),  or  n  =  1,  which  gives  generally,  <£  —  -,  when  the 

angle  <f>  is  expressed  in  radians. 

The   preceding   demonstration  shows   that   the   radian   is 

180° 
invariable,  and  is  equivalent  to  -  — ,  or  57°.2957795+ .     The 

equation,  <j>  =  -  ,  r  =  -  ,  or  s  —  r<^>,  is  of  practical  importance 
r  <f> 

according  as  <£,  r,  or  s,  respectively,  is  the  unknown. 

If  8  represents  the  number  of  degrees,  y  the  number  of 
grades,  and  p  the  number  of  radians  in  an  angle,  we  have,  as 
in  the  preceding  demonstration  : 

_8_        y  s 

360  ~~  400  ~~  2  ?rr '  C 

-L===JL.S=£. 

180      200      TT 

From  which  follow  these  six  equations  of  transformation : 

8  =  JL        =??5       -    *•  - 


10  TT  180 


LINEAR  AND  ANGULAR  MAGNITUDE. 


Making  the  symbols  8,  y,  p  in  the  right  members  of  these 
equations  successively  equal  to  unity,  we  obtain  the  value  of 
each  unit  in  terms  of  the  others.  This  gives  the  following 
table  : 


In  terms  of 

Units 

o 

r 

g 

7t 

10 

1° 

1 

180 

9 

180 

200 

lr 

1 

rt 

Tt 

9 

it 

l^ 

10 

200 

1 

Degree  measure  is  used  in  practical,  and  circular  measure 
in  theoretical  work. 

c.   Examples.    I. 

1.  Express  2523°  17'  32"  in  grades,  and  radians. 

2.  Express  57s  17'  94"  in  degrees,  and  radians. 

3.  Express  2.3r  in  degrees,  and  grades. 

,    „  tfr    ?rr    TC*    . 

4.  Express  TV,  —  ,  — ,  — ,  in  degrees. 

5.  Assuming  the  earth  to  be  a  sphere  with  a  radius  of  3963  miles, 
compute  the  length  of  a  geographical  mile  in  statute  miles. 

6.  Find  the  distance,  measured  on  the  arc  of  a  great  circle,  in  miles, 
from  Oberlin  (lat.  41°  17'  N.)  to  the  equator. 

7.  A  man  stands  on  the  extremity  of  a  radius  of  a  circular  platform 
which  is  supported  on  a  vertical  axis  through  its  center,   and  perpen- 
dicular to  its  plane.     The  extremity  of  the  given  radius  is  opposite  a 
fixed  arrowhead.    The  platform  is  made  to  rotate  about  its  axis  counter- 
clockwise, and  finally  comes  to  rest.     The  man  has  moved  through  a 
distance  of  £  mi.     Find  in  degrees  the  actual  angle  through  which  the 
platform  has  turned,  and  the  apparent  angle  which  the  given  radius 
makes  with  the  line  of  the  fixed  arrow,  the  radius  being  15  ft.  in  length. 


CHAPTER   II. 


A.   Conventions. 
a.   Conventions  about  Lines. 

§  9.  A  plane  is  divided  into  four  quadrants  by  two  fixed 
perpendicular  straight  lines.  The  plane  may  always  be  so 
placed  that  one  line  is  horizontal,  and  the  other  vertical.  For 
convenience  we  shall  speak  of  the  lines  as  the  horizontal,  and 
the  vertical  line,  or  axis. 

§  1O.  The  point  of  intersection  of  the  two  lines  is  called 
the  origin,  and  is  denoted  by  the  letter  0.  The  horizontal 
axis  is  called  the  x-axis,  the  vertical,  the  y-axis. 

§  11.  The  distance  of  any  point  measured  from  the  y-axis 
parallel  to  the  x-axis,  will  be  represented  by  x,  and  called  its 
abscissa ;  and  the  distance  of  any  point  measured  from  the 
x-axis  parallel  to  the  y-axis,  will  be  represented  by  y,  and 
called  its  ordinate.  The  distance  of  any  point  from  the 
origin  will  be  represented  by  r,  and  called  its  radius  vector. 


x  =  OM. 
y  =  M  P.  X- 
r  =  OP. 


0 


Y' 
FIG.  2. 


TRIANGLE  OF  REFERENCE.  7 

§  12.  We  will  adopt  the  convention,  that  is,  the  AGREEMENT 
that  the  direction  FROM  left  TO  right  of  FIXED  horizontal  lines 
is  POSITIVE,  the  OPPOSITE  direction  is  NEGATIVE  ;  and  that  the 
direction  FROM  below  UPWARDS  of  FIXED  vertical  lines  is 
POSITIVE,  the  OPPOSITE  direction  is  NEGATIVE. 

§  13.  By  convention,  the  directions  of  the  sides  of  an  angle 
are  said  to  be  positive  if  taken  from  the  vertex  outwards,  the 
opposite  directions  are  negative. 

b.  Conventions  about  Angles. 

§  14.  The  turning  of  a  line  which  generates  an  angle  may 
be  either  counter-clockwise  or  clockwise.  Counter-clockwise 
rotation  we  will  call,  by  convention,  positive;  the  opposite, 
negative.  An  angle  generated  by  positive  rotation  we  will  call 
positive  ;  an  angle  generated  by  negative  rotation,  negative. 

§  15.  The  four  quadrants  are  numbered  the  first,  the 
second,  the  third,  and  the  fourth,  respectively,  beginning 
with  XOY,  and  taken  in  the  order  of  positive  rotation. 

§  16.  The  initial  line  of  any  single  angle  will  always  be 
taken  as  coinciding  with  OX,  and  the  angle  will  be  said  to  be 
of  that  quadrant  in  which  its  terminal  line  lies.  In  any 
composite  angle  the  initial  line  of  any  angle  coincides  with 
the  terminal  line  of  the  angle  preceding,  and  may,  for  the 
moment,  be  considered  as  coinciding  with  OX. 

B.  Triangle  of  Reference. 

§  17.  If  at  any  point  of  the  x-axis,  or  of  the  y-axis,  a  per- 
pendicular be  erected  and  produced  to  intersect  the  rotating  line 
of  a  given  angle  at  the  point  P,  a  right  triangle  will  be  formed, 
which  is  called  the  TRIANGLE  OF  REFERENCE  for  the  given 
angle.  The  abscissa  of  P  is  always  the  base,  its  ordinate,  the 
perpendicular,  and  its  radius  vector,  the  hypotenuse  of.  the 
triangle  of  reference. 


PLANE  TRIGONOMETRY. 


C.   Trigonometric  Functions. 

§  18.  The  six  ratios  which  can  be  formed  by  using  the  three 
sides  of  the  triangle  of  reference  of  a  given  angle,  two  at  a  time, 
are  called  the  six  primary  trigonometric  functions  of  the  angle. 

The  sine  of  the  angle  is  the  ratio  of  the  perpendicular  to  the 
hypotenuse. 

The  cosine  of  the  angle  is  the  ratio  of  the  base  to  the  hypote- 
nuse. 

The  tangent  of  the  angle  is  the  ratio  of  the  perpendicular  to 
the  base. 

The  cotangent  of  the  angle  is  the  ratio  of  the  base  to  the  per- 
pendicular. 

The  secant  of  the  angle  is  the  ratio  of  the  hypotenuse  to  the 
base. 

The  cosecant  of  the  angle  is  the  ratio  of  the  hypotenuse  to  the 
perpendicular. 

§  19.  The  following  figures  show  to  the  eye,  how,  by  the 
definition,  triangles  of  reference  are  constructed  for  an  angle 
of  each  quadrant. 


X- 


Y' 
FIG.  3. 


x       M 


TRIGONOMETRIC  FUNCTIONS. 
Y 


Y' 

FIG.  5. 


§  2O.  It  is  to  be  observed,  and  the  student  should  carefully 
prove,  that  the  triangle  of  reference  for  any  angle  includes 
four  species,  that  all  the  triangles  that  can  be  constructed 
are  geometrically  similar,  and  that  in  passing  from  a  triangle 
of  one  species  to  a  triangle  of  any  other  species,  either  no 
change,  or  two  changes  of  sign  occur  in  the  terms  of  any 
trigonometric  function,  and  that,  therefore,  the  function  is 
unaltered.  This  work  should  be  done  as  is  shown  below  for 
the  sine  and  cosine  of  an  angle  of  the  first  quadrant. 


10 


PLANE  TRIGONOMETRY. 


perpendicular 

sine  of  <£,  written  sin  <£,  =  *-r—J          --  = 

hypotenuse 


r2       OP3 
cosine  of  <£,  written  cos  <£,  = 


r3       OP4 

base 
hypotenuse      0 


OMt 


M2P 

—  2— 

OP2 


OM2 
OPS 


__  X2  _   M3P8  _  Xg  _  M4P4  _  X4  ^ 

~  r2~  OP3  ~r3        OP4  ~  r4' 

We  shall  use  as  abbreviations  for  tangent  of  <£,  cotangent 
of  <£,  secant  of  ^,  and  cosecant  of  <j>,  respectively,  tan  <£, 
ctn  <^>,  sec  <£,  and  esc  <^>. 

§  21.  Since  the  trigonometric  functions  are  abstract  num- 
bers, all  the  arithmetical  operations  can  be  performed  upon 
them.  It  is  customary  to  indicate  the  nth  power  of  sin  <f> 
by  sinn  <f> ;  similarly  for  the  other  trigonometric  functions. 

§  22.  In  addition  to  the  primary  functions  previously  de- 
fined, two  secondary  functions  are  usually  defined  as  follows : 

The  versed-sine  of  <£,  written  vrs  <£,    =1  —  cos  <j>, 
The  coversed-sine  of  <£,  written  cvs  </»,  =  1  —  sin  <£. 


MUTUAL  RELATIONS  OF  FUNCTIONS. 


11 


Two  others  are  more  rarely  defined  as  follows  : 

The  suversed-sine  of  <j>,  written  svs  <f>,   =  (2  —  vrs  <£), 

—  1  +  cos  <£. 
The  sucoversed-sine  of  <£,  written  scs  <£,  =  (2  —  cvs  <£), 

=  1  +  sin  <£. 

§  23.  By  referring  to  Figs.  3-6  we  see  that  the  sine, 
cosine,  secant,  and  cosecant  of  an  angle  of  the  third  quadrant 
are  negative,  because  the  terms  of  these  ratios  have  opposite 
signs,  and  that  its  tangent  and  cotangent  are  positive,  because 
the  terms  of  these  ratios  have  the  same  sign.  Similarly,  by 
a  careful  examination  of  Figs.  3-6,  the  student  should  prove 
the  following  table  for  the  signs  of  the  functions  of  angles 
of  the  four  quadrants. 


Functions. 

Quadrants. 

1 

2 

3 

4 

sine  and  cosecant  .... 

+ 

+ 

- 

- 

cosine  and  secant  .... 

+ 

- 

- 

+ 

tangent  and  cotangent    . 

+ 

- 

+ 

- 

Z>.  Mutual  Relations  of  Functions. 

§  24.    From  the  definitions  of  the  trigonometric  functions 
it  is  evident  that : 


sn   >  = 


CSC 


cos  <   = 


sec 


;  tan  <£  = 


1 


ctn 


that  is,  the  sine  and  the  cosecant,  the  cosine  and  the  secant, 
the  tangent  and  the  cotangent  of  an  angle  <f>,  are  the  recip- 
rocals of  each  other,  always. 


12  PLANE   TRIGONOMETRY. 

sin  (f>  cos  <£ 

0.  tan  <i  = :  ;  ctn  6  =  — : >  always. 

cos  <j!> '  sin  <£ 

0.  Since,  of  whatever  quadrant  the  angle  <f>  may  be,  that  is, 
whatever  may  be  the  intrinsic  signs  of  x,  y,  and  r,  x2,  y2,  and  r2 
are  always  positive,  and  y2  +  x2  =  r2,  always, 

y2      x2 
.'.—-] — -  =  1,     that  is,  sin2</>  +  cos2</>  =  1,  always. 

v2  r2 

Similarly,  Jg  + 1  =  -5 ,  that  is,  tan2<£  +  1  =  sec'2<£,  always. 


r2 

> 


x  r 

And  —^  +  1  =  1  >  that  is,  ctn2</>  -j-  1  =  csc2^>,  always. 


d.  From  the  preceding  relations,  it  is  evident  that  some  of 
the  functions  may  be  expressed  in  terms  of  others.  It  can 
be  shown  that  all  the  functions  may  be  expressed  in  terms  of 
any  one,  which  may  be  chosen  at  pleasure  as  the  independent 
function.  For  example,  all  the  functions  may  be  expressed 
in  terms  of  the  tangent  as  follows  : 

ctn  cf>  — •     Since  tan2</>  +  1  =  sec2<£, 

tan  <p 

.'.  sec  <£  =  ± 

Since  cos  <f>  = ,  .*.  cos  <i  =  ± 

sec  </> 

sin  <f>  .        _ 

cos  </> ' 

1  Vtan  V  +  1 

esc  <f>  =  — ,  .'.  csc  <j>  =  ±  — 

sin  <£  tan  <£ 

It  is  left  as  an  exercise  for  the  student  to  complete  the 
following  table. 


1 

Since 

Vtan2</>  -j-  1 

tan  ^> 

Vtan2<^>  -|~  1 

EXAMPLES.    II. 


13 


Func- 
tions. 

In  terms  of 

sin  0 

COS  0 

tan  0 

ctn  0 

sec  0 

CSC  0 

sin  0 

.         tan0 

"~  Vtan20  +  1 

COS  0 

±           1 

^  Vtan20  +  1 

tan  0 

tan0 

ctn  0 

1 

tan0 

sec  0 

±  Vtan20  +  1 

CSC  0 

±  Vtan20  +  1 

tan  0 

E.  Examples.    II. 


1.  Of  what  quadrant  is  each  of  the  following  angles  ?  38°,  317°,  2530°, 
-  1155°,  275s,  31',  -  13.4',  -823°,  -423?,  625°  43'  26",  761  e  87'  95", 
-27.364'. 

2.  Determine  the  sign  of  each  of  the  trigonometric  functions  of  the 
angles  in  example  1. 

3.  If  0  is  an  angle  of  the  first  quadrant,  find  the  values  of  the  other 
trigonometric  functions,  having  given  :  sin  0  =  |,  cos  0  =  if,  tan  0  =  ^ 
ctn  0  =  f ,  sec  0  =  3,  esc  0  =  |. 

4.  If  0  is  an  angle  of  the  second  quadrant,  find  the  other  trigonometric 
functions,  having  given  :   sin  0  =  |,  cos  0  =  —  f ,  tan  0  =  —  |,   ctn  0  = 
—  2,  sec  0  =  —  7,  esc  0  =  10. 

5.  If  0  is  an  angle  of  the  third  quadrant,  find  the  other  trigonometric 
functions,  having  given  :  sin  0  =  —  -£,  cos  0  =  —  V^,  tan  0  =  VJJ,  sec  0 
=  —  5,  ctn  0=1,  esc  0  =  —  2. 

6.  If  0  is  an  angle  of  the  fourth  quadrant,  find  the  other  trigonometric 

V3-1 


functions,  having  given  :  sin  0  =  — 


ctn  0  =  —  2  V6,  sec  0  = 


2V2 
V3  +  1 


2V2    ' 


=  1,  tan0  =  —  V8, 


,    CSC  0  =  — 


14  PLANE   TRIGONOMETRY. 

7.  Determine  all  the  possible  values  of  the  other  trigonometric  func- 

tions,  having  given  :    sin  0  =  —  V|,    cos  0  =  —  Vfc,    tan  0  =  —  -  » 
ctn  0  =  —  4  V3,  sec  0  =  V2,  esc  0  =  a. 

8.  If  0  is  an  angle  of  any  quadrant,  draw  triangles  of  reference  for 

the   following  angles  :    4  n  |  +  0,    (4  n  +  1)  |  +  0,    (4  n  +  2)  ^  -f  0, 


Prove  the  following  identities  : 

9.  sin  0  esc  0  =  cos  0  sec  0  =  tan  0  ctn  0  =  1. 

10.  cos  0  +  vrs  0  =  sin  0  +  cvs  0  =  svs  0  —  cos  0  =  scs  0  —  sin  0  =  1. 

11.  vrs  0  svs  0=1  —  cos20  =  sin20. 

12.  cvs  0  scs  0  =  1  —  sin20  —  cos20. 

13.  svs  0  scs  0  =  1  +  sin  0  +  cos  0  +  sin  0  cos  0. 

14.  cos  0  tan  0  =  sin  0. 

15.  sin  0  ctn  0  =  cos  0. 

16.  (sec  0  +  tan  0)  (sec  0  —  tan  0)  =  (esc  0  +  ctn  0)  (esc  0  —  ctn  0)  =  1. 

17.  (sin  0  +  cos  0)2  +  (sin  0  —  cos  0)2  =  2. 

18.  tan  0  +  ctn  0  =  sec  0  esc  0. 

19.  cos20  —  sin20  =  2  cos20  —1  =  1  —  2  sin20. 

20.  (cos  0  +  sin  0)  (esc  0  —  sec  0)  =  ctn  0  —  tan  0. 

21.  sin  0  (1  +  tan  0)  +  cos  0  (1  +  ctn  0)  =  sec  0  +  esc  0. 

22.  sin40  +  cos40  =  1—2  sin20  cos20. 

23.  sin40  —  cos40  =  sin20  —  cos20. 

24.  tan20  —  ctn20  =  sec20  —  csc20. 

25.  (1  +  tan  0)2  +  (1  -  tan  0)2  =  2  sec20. 

26.  (1  +  cos  0)2  +  (1  +  sin  0)2  =  3  +  2  (sin  0  +  cos  0). 

27.  (sin  0  +  cos  0)  (tan  0  +  ctn  0)  =  sec  0  +  esc  0. 

28.  (sec  0  +  esc  0)2  =  (1  +  tan  0)2  +  (1  +  ctn  0)2. 

29.  tan  0  (1  —  ctn2  0)  +  ctn  0  (1  —  tan2  0)  =  0. 

30.  sin30  +  cos30  =  (sin  0  +  cos  0)  (1  —  sin  0  cos  0). 

31.  sin30  —  cos30  =  (sin  0  —  cos  0)  (1  +  sin  0  cos  0). 

32.  (.«,,  -•*.,)»=  -=*  • 


33. 

34.  esc  0  +  ctn  0  = 


esc  0  —  ctn  0 
35. 


EXAMPLES.    II.  15 


36. 
37 


ctn  0  +  ctn  e 
ctn4.  +  tanfl  =  ct 
tan  0  +  ctn  0 


tan  0  +  ctn  0 
39.  sin20  tan20  +  cos20  ctn"20  =  tan20  +  ctn20 

sin  0  +  sin  0  _  cos  0  +  cos  d 
'  cos  0  —  cos  6  ~  sin  0  —  sin  0 

,    tan  <j>  +  sec  0  —  1 
41.    —  -  -  -  -  =  tan  0  +  sec  0. 
1  +  tan  0  —  sec  0 

1  —  sec  0  +  tan  0  _  sec  0  +  tan  0  —  1  < 
1  +  sec  0  —  tan  0  ~~  sec  0  +  tan  0  +  1 
1  +  esc  0  +  ctn  0  _  esc  0  +  ctn  0  —  1  ^ 
1  +  esc  0  —  ctn  0      ctn  0  —  esc  0  +  1 
44.  (1  4-  sec  0  +  tan  0)  (1  +  esc  0  +  ctn  0) 
=  2  (1  +  tan  0  +  ctn  0  +  sec  0  +  esc  0). 


CHAPTEK   III. 

A.   Variation  of  Trigonometric  Functions. 

§  25.  It  is  necessary  to  trace  the  changes  in  the  values  of 
the  trigonometric  functions  as  the  angle  changes.  The  fol- 
lowing two  examples  are  sufficient  to  show  how  this  is  done. 

Beginning  with  an  indefinitely  small  angle,  we  will  trace 
the  changes  in  the  sine  and  tangent  of  the  angle  as  the  angle 
increases.  One  of  the  triangles  of  reference  which  lie  in  the 
same  quadrant  as  the  terminal  line  of  the  angle,  will  be  used. 
In  the  discussion  of  the  sine,  the  hypotenuse  of  the  triangle 
of  reference  will  be  kept  constant  in  length.  The  sign  ~ 
will  be  used  for  the  words  "  approaches  as  its  limit."  0°  is 
the  limit  of  an  indefinitely  small  angle.  It  is  evident  that 
the  sine  of  an  indefinitely  small  angle  must  be  indefinitely 
small,  since  the  perpendicular  of  the  triangle  of  reference 
must  be  indefinitely  small ;  that  is,  sin  a°  =  e,  where  a°  =  0°, 
and  e  =  0.  Since  the  variables  sin  a°  and  e  are  always  equal 
however  near  they  approach  their  limits,  therefore  their 
limits,  sin  0°  and  0,  are  equal  ;  that  is,  sin  0°  =  0.  As  the 
angle  increases  between  0°  and  90°,  the  perpendicular  of  the 
triangle  of  reference  increases,  therefore  the  sine  of  the  angle 
increases,  and  vice  versa.  When  an  angle  approaches  90°  as 
its  limit,  the  length  of  the  perpendicular  approaches  as  its 
limit  the  length  of  the  hypotenuse  ;  therefore,  as  in  the  pre- 
ceding case,  sin  90°  =  1.  As  the  angle  increases  between  90° 
and  180°,  the  perpendicular  decreases,  therefore  the  sine  of 
the  angle  decreases,  and  vice  versa. 

When  an  angle  approaches  180°  as  its  limit,  its  sine  ap- 
proaches 0  as  its  limit,  and  sin  180°  =  0.  The  sine  of  an 
angle  in  the  third  and  fourth  quadrants  is  negative.  As  $he 


VARIATION  OF  TRIGONOMETRIC  FUNCTIONS.  17 

angle  increases  between  180°  and  270°,  the  perpendicular 
increases  in  length,  that  is,  decreases  in  algebraic  value, 
therefore  the  sine  decreases,  and  vice  versa.  By  the  use  of 
limits  it  is  evident  that  sin  270°  =  —  1.  As  the  angle  in- 
creases between  270°  and  360°,  the  perpendicular  decreases 
in  length,  that  is,  increases  in  algebraic  value,  therefore  the 
sine  increases,  and  vice  versa.  And  by  the  use  of  limits 
sin  360°  =  sin  0°  =  0.  If  the  angle  should  continue  to  in- 
crease, the  same  set  of  values  of  the  sine  would  recur  period- 
ically at  intervals  of  360°.  This  recurrence  of  values  is 
called  the  periodicity  of  the  sine,  and  will  be  noticed  again. 

§  26.  In  the  discussion  of  the  tangent,  the  base  of  the 
triangle  of  reference  will  be  kept  constant  in  length.  By 
reasoning  like  that  used  for  the  sine  it  is  evident  that 
tan  0°  =  0,  and  that  as  the  angle  increases  between  0°  and 
90°  the  tangent  increases,  and  vice  versa.  As  an  angle  ap- 
proaches 90°  as  its  limit,  the  perpendicular  of  the  triangle  of 
reference  increases  indefinitely,  therefore  the  tangent  in- 
creases in  numerical  value  without  limit,  which  is  what  is 
meant  when  we  say  tan  90°  —  oo  .  In  the  second  quadrant 
the  base  of  the  triangle  of  reference  is  negative,  hence  the 
tangent  is  negative.  As  the  angle  increases  between  90°  and 
180°,  the  perpendicular  decreases,  therefore  the  tangent  in- 
creases in  algebraic  value,  and  vice  versa.  And  by  the  doc- 
trine of  limits  tan  180°  =  0.  In  the  third  quadrant  both 
the  base  and  the  perpendicular  of  the  triangle  of  reference 
are  negative,  hence  the  tangent  is  positive.  As  the  angle 
increases  between  180°  and  270°,  the  perpendicular  increases, 
therefore  the  tangent  increases,  and  vice  versa.  As  the  angle 
approaches  270°  as  its  limit,  the  perpendicular  increases 
indefinitely,  therefore  the  tangent  increases  in  numerical 
value  without  limit,  and  we  say,  as  before,  tan  270°  =  oo  . 
In  the  fourth  quadrant,  the  perpendicular  only  of  the  triangle 
of  reference  is  negative,  hence  the  tangent  is  negative.  As 


18 


PLANE  TRIGONOMETRY. 


the  angle  increases  between  270°  and  360°,  the  perpendicular 
decreases  in  length,  therefore  the  tangent  increases  in  alge- 
braic value,  and  vice  versa.  And  by  the  doctrine  of  limits 
tan  360°  =  tan  0°  =  0.  The  tangent  also  has  a  recurrence 
of  values,  at  intervals  of  180°. 

§  27.   The  following  pictures,  called  graphs,  of  the  pre- 


f 


7 

Y' 
FIG.  8.  —  Graph  of  the  Sine. 

Y 


FIG.  9.  — Graph  of  the  Tangent. 


ceding  tracing  of  values  are  given  to  aid  the  imagination. 
Abscissas  represent  values  of  the  angle  expressed  in  circular 
measure ;  ordinates  represent  the  corresponding  values  of 
the  functions. 


VARIATION  OF  TRIGONOMETRIC  FUNCTIONS. 


19 


The  student  should  now  trace  the  changes  in  the  values  of 
the  other  trigonometric  functions,  and,  independently,  obtain 
a  table  like  the  following. 


Angle 

sin 

cos 

tan 

ctn 

sec 

CSC 

0° 

TO; 

i 

TO 

=j=co 

1 

=Fco 

-360° 

Qi 

inc 
dec 

dec 
inc 

inc 
dec 

dec 
inc 

inc 
dec 

dec 
inc 

Qi 

90° 

I 

±0 

±00 

±0 

±00 

1 

—270° 

Q2 

dec 

inc 

dec 
inc 

inc 
dec 

dec 
inc 

inc 
dec 

inc 
dec 

Q2 

180° 

±0 

-1 

TO 

qpoo 

-1 

±00 

-180° 

Q3 

dec 
inc 

inc 
dec 

inc 
dec 

dec 
inc 

dec 
inc 

inc 
dec 

Q3 

270° 

-1 

TO 

±00 

±0 

T°° 

-1 

-90° 

Q4 

inc 

dec 

inc 
dec 

inc 
dec 

dec 
inc 

dec 

inc 

dec 
inc 

Q4 

360° 

=1=0. 

1 

TO 

=FCO 

1 

=poo 

0° 

sin 

cos 

tan 

ctn 

sec 

CSC 

Angle 

In  this  table,  for  an  angle  generated  by  positive  rotation, 
i.e.,  an  increasing  angle,  the  upper  word  goes  with  the  value 
of  the  angle  taken  from  the  left-hand  column  ;  and  for  an 
angle  generated  by  negative  rotation,  the  lower  word  goes 
with  the  value  of  the  angle  taken  from  the  right-hand  column. 
The  -f~  and.  —  before  0  and  oo  indicate  how  the  function 
changes  sign  as  the  angle  passes  through  the  quadrant  limit. 


20 


PLANE   TRIGONOMETRY, 


B.  Line  Representations  of  the  Trigonometric 
Functions. 

§  28.  If  in  each,  of  the  primary  trigonometric  functions, 
or  ratios,  a  positive  linear  unit  is  taken  for  the  denominator, 
the  number  of  linear  units  in  the  numerator  will  represent 
the  abstract  number  of  units  in  the  ratio,  and  the  line  which 
is  the  numerator  of  the  ratio  will  completely  represent  the 
function.  The  student  should  remember  that  the  line  is  not 
the  function,  but  represents  it.  This  restriction  gives  the 
following  constructions. 


REPRESENTATIONS  OF  TRIGONOMETRIC  FUNCTIONS.    21 


MP 
In  these  figures,  for  example,  sin  <f>  =  - -- ,  but  OP  is  taken 

equal  to  a  linear  unit.  Hence,  sin  <£  is  represented  by  M  P. 
cos  (f>  is  represented  by  OM.  Similarly  the  student  should 
prove  for  each  quadrant,  that 

tan  <f>  is  represented  by  M'P'? 
ctn  <f>  is  represented  by  N  Q, 
sec  (j>  is  represented  by  OP', 
esc  <f>  is  represented  by  OQ. 

The  secondary  trigonometric  functions  are  also  represented 
by  lines  as  follows  : 

vrs  <j>  is  represented  by  MM', 
cvs  <j>  is  represented  by  LN, 
svs  <f>  is  represented  by  R  M, 
scs  <j>  is  represented  by  S  L. 

§  29.  The  student  should  ascertain  the  signs  of  the  func- 
tions in  the  four  quadrants  by  means  of  the  line  representa- 
tives. The  accompanying  figure 
is  intended  forcibly  to  impress 
on  the  student's  mind  that  the 
sine  and  cosecant  of  an  angle 
of  a  quadrant  above  the  hori- 
zontal line  are  positive,  below 
negative  ;  that  the  cosine  and 
secant  of  an  angle  to  the  right 
of  the  vertical  line  are  positive, 
to  the  left  negative  ;  and  that 
the  signs  of  the  tangent  and 
cotangent  are  given  by  the 
combination  of  the  signs  of  the 


FIG.  14. 


sine  and  cosine,  quadrant  by  quadrant, 
seen  by  line  representation. 


This  is  all  readily 


22  PLANE   TRIGONOMETRY. 

C.  Inverse  Trigonometric  Functions. 

§  3O.  If  two  variables  are  connected  by  any  relation,  either 
one  of  them  may  be  considered  as  the  independent  variable, 
and  the  other  as  the  dependent  variable  or  direct  function. 
When  the  dependent  variable  is  made  independent,  the  in- 
dependent variable  becomes  the  dependent  variable  and 
inverse  function. 

If  x  is  any  function  of  the  angle  <f>,  </>  is  the  angle  whose 
function  is  x.  These  two  statements  of  the  same  relation, 
viewed  from  the  different  standpoints  alluded  to,  are  symbol- 
ized as  follows  :  x  =  f  (<£),  and  <£  =  f"1  (x),  and  are  read,  x  is 
a  function  of  <£,  and  <j>  is  the  anti-function  of  x.  For  example, 
if  x  =  sin  <£,  <f>  is  the  angle  whose  sine  is  x,  which  is  written 
<f>  =  sin^x,  and  is  read,  <j>  is  the  anti-sine  of  x.  Some  writers 
express  the  inverse  functional  relations  as  follows  :  <f>  =  arc 
sin  x,  0  =  arc  cos  x,  and  similarly  for  the  other  functions. 

/>.   Examples.    III. 

1.  By  representing  positive  numbers  by  distances  to  the  right  of  the 
origin  on  the  x-axis,  and  negative  numbers  by  distances  to  the  left  of 
the  origin,  and  the  values  which  functions  can  have  by  heavy  lines,  prove 
that  the  following  is  a  correct  graphic  representation  of  the  possible 
range  of  values  of  the  primary  trigonometric  functions. 


sin  and  cos  X 
sec  and  esc  X 
tan  and  ctn  X'1 


—  oo 

0                                              +00 

-1 

+  1 

—  oo 

0                                  +00 

—  1 

+  1 

—  oo 

0                                 +00 

V 

EXAMPLES.    III. 


23 


The  results  of  this  problem  are  very  important.  The  student  should 
carefully  remember  both  the  range  of  possible  and  impossible  values  of 
the  functions. 

Trace  the  changes  in  value  of  the  following  expressions  : 

6.  vrs  0.  8.  svs  0. 


2.  cos  0. 

3.  ctn  0. 


4.  sec  0. 

5.  esc  0. 


7.  cvs  0. 


9.  scs  0. 


Construct  an  angle  : 

10.  (a)  Of  the  1st,  and  of  the    2d 
"  4th 

"  4th 

3d 
3d 

"  4th 

3d 
4th 

"  4th 

3d 

"  2d 

"  4th 


(b)      « 

3d 

11. 

(a)      " 

1st 

(b)      « 

2d 

12. 

(a)      « 

1st 

(b)      " 

2d 

13. 

(a)      " 

1st 

(b)      " 

2d 

14. 

(a)      " 

1st 

(6)      " 

2d 

15. 

(a)      " 

1st 

(6)      " 

3d 

quadrant,  whose  sine  is  £. 

U  t(  It  1 

"    cosine  is  i. 

-±. 

".  .         "  tangent  is  2. 

it  i i .        a,          o 

"      ctn    is*. 


secant  is  V2. 

-V2. 
esc    is  V5.  _ 

"       -Vs. 


Construct  all  possible  values  of  : 

16.  sin-1  (±  f ).  18.  tan-1  (±  3).  20.  sec-1  (±  5). 

17.  cos-1(±S).  19.  ctn-^if).  21.  esc-1  (±  2). 

22.  Can  the  following  be  constructed?    (a)  sin-1!,    (&)  cos-1  (—  |), 
(c)  tan-1  m,  (d)  ctn-1  (±  676),  (e)  sec-1  (db  i),  (/)  esc-1  VS. 

23.  Are  the  following  possible  ? 

(a)  tan-1  (±  36,947),    (b)  sec-1  (±  0.7435),    (c)  sin-1  (±  25). 


CHAPTER   IV. 

A.   Trigonometric  Functions  of  Integral  Multiples 

of  |>  ±<f>. 

§  31.  The  complement  of  an  angle  is  the  angle  obtained 
by  subtracting  the  given  angle  from  90°.  The  supplement 
of  an  angle  is  the  angle  obtained  by  subtracting  the  given 
angle  from  180°.  Thus,  the  complement  of  50°  is  40°;  the 
complement  of  125°  is  —  35°.  The  supplement  of  110°  .is 
70°;  the  supplement  of  200°  is  —20°. 

Two   cases   of  integral  multiples  of  —  are  to  be  distin- 

a 

guished  ;  first,  when  the  multiplier  is  odd  ;  second,  when  it 

is  even.     Under  the  first  case  the  simplest  multiples  of  — 

& 

are,  when  the  angle  is  expressed  in  degrees,  90°  and  270°, 
the  multipliers  being  1  and  3,  respectively.  Under  the 
second  case  the  simplest  examples  are,  similarly,  0°,  180°, 
and  360°,  the  multipliers  being  0,  2,  and  4,  respectively. 
The  examples  enumerated  lead  us  to  seek  the  trigonometric 
functions  of  the  special  angles  ±  <£,  (90°±  </>),  (180°db  <£), 
(270°±  <£),  and  (360°±  <j>)  in  terms  of  the  functions  of  <j>. 

In  comparing  the  functions  of  (m  90°±  <£)  with  the  func- 
tions of  <£  the  triangles  of  reference  will,  for  convenience, 
be  made  geometrically  equal  by  keeping  the  hypotenuse  con- 
stant in  length  ;  but  it  is  to  be  observed  that  the  demonstra- 
tions in  no  wise  require  this  restriction,  but  depend  on  the 
underlying  principle  that  the  homologous  sides  of  similar 
triangles  are  proportional.  Also  in  this  discussion  only  the 
primary  functions  will  be  considered. 


INTEGRAL  MULTIPLES. 


25 


§  32.  The  functions  of  <£  have  already  been  given.  If  <j>  is 
taken  in  any  quadrant,  the  terminal  lines  of  <j>  and  of  ( —  <£) 
will  lie  on  the  same  side  of  the  y-axis  and  on  opposite  sides 
of  the  x-axis.  From  the  numerical  equality  of  the  angles,  the 
triangles  of  reference  for  </>  and  for  ( —  <£)  can  always  be  con- 
structed geometrically  equal,  by  drawing  a  single  perpendicular 
to  the  x-axis  and  producing  it  in  opposite  directions  until  it 
cuts  the  terminal  lines.  Trigonometrically  the  bases  and 
the  hypotenuses  of  the  triangles  of  reference  are  equal,  the 
perpendiculars  are  negatives  of  each  other.  It  follows  that 
all  the  functions  of  ( —  <£)  into  which  the  perpendicular  of  the 
triangle  of  reference  enters,  will  be  the  negatives  of  the  same 
functions  of  </>,  and  that  the  others  will  be  equal  to  the  same 
functions  of  <£. 

The  following  figure  illustrates  the  preceding  results. 


Y' 


FIG.  15. 


We  have  here  ^  =  x,  rl  =  r,  and  yr  =  —  y.     Therefore, 

sin  (—  0)  =  ^=:^  =— sin<£,  esc  (—  <£)  =  —  =  --=— csc  <£. 
i"i          r  YI  Y 

COS(—  <f>)  =  *l  =     '-'-     =       COS<£,  Sec  (—<£)  =  -=:     ;-     =  S6C  <£. 

i"i          r  Xi 


26 


PLANE   TRIGONOMETRY. 


§  33.    In  constructing  the  triangle  of  reference  for  the  angle 
(m  90°  +  <£),  the  angle  will  be  constructed  as  (<j>  +  m  90°). 
We  will  first  consider  the  cases  where  m  =  0,  2,  and  4. 
<f>  +  0.90°  =  <f>,  whose  functions  have  been  given  by  definition. 

If  180°  is  added  to  <£  in  any  quadrant,  the  triangles  of 
reference  for  (<£  +  180°)  and  <£  are  geometrically  equal.  The 
hypotenuses  are  equal,  and  the  perpendiculars  and  bases  are 
respectively  negatives  of  each  other.  It  follows  that  those 
functions  of  (<£  +  180°)  into  which  either  the  base,  or  the 
perpendicular,  enters  singly,  are  the  negatives  of  the  same 
functions  of  <£ ;  and  that  those  functions  of  (<j>  -+- 180°)  into 
which  both  the  base  and  the  perpendicular  enter  are  equal  to 
the  same  functions  of  <f>. 

The  following  figure  illustrates  the  preceding  results. 


We  have  here  r1  =  r,  xl  =  —  x,  and  yx  =  —  y.     Therefore, 

sin  (180°  +  <£)  =  ^  =  ^  = 
TI 

esc  (180°  - 


—  sin 


=  —  esc  d>, 


=  —  = =  —  cos 


sec  (180°  +  <h)  =  —  =  —  =  —  sec  <f>. 

\  r/  


INTEGRAL  MULTIPLES.  27 

tan (180°  +$)  =  £=—-2  =  tan  <j>, 
Xj      —  x 

ctn  (180°  +  <#>)  =  -  =  ^  -  =  ctn  <£. 

It  is  to  be  observed  that  the  triangle  of  reference  for 
(<£  +  180°)  may  be  obtained  by  revolving  the  triangle  of 
reference  for  <£  about  the  origin  in  the  plane  of  the  figure 
through  a  positive  angle  of  180°.  This  reverses  the  direc- 
tions, and,  therefore,  the  signs  of  the  base  and  the  perpen- 
dicular. 

It  is  also  to  be  observed  that  in  the  preceding  discussion 
<£  is  an  angle  of  any  quadrant,  therefore  is  unrestricted,  and 
may  be  positive  or  negative. 

§  34.  If  <£  is  replaced  by  (—  <£),  the  formulas  "of  §  33 
become,  by  the  use  of  the  formulas  of  §  32, 

sin  (180°  —  $)  =  —  sin  (—  <£)  =  sin  <j>, 
.'.esc  (180°  —  <£)=csc<£. 

cos  (180°  —  <£)  =  —  cos  (—<£)=  —  cos  <£, 
/.sec  (180°  —  <£)  =  —  sec  <£. 

tan  (180°  —  <£)  =  tan  (—  <£)  =  —  tan  <£, 
.\ctn  (180°  —  </>)=  —  ctn  <j>. 

Observing  that  (<£  —  180°)  =  —  (180°  —  <£),  we  have  by  the 
formulas  of  §  32,  and  the  present  section, 

sin  (<£  —  180°)  =  —  sin  (180°  —  <£)  =  —  sin  <£, 
.'.  esc  (<£  —  180°)  =  —  esc  <£. 

Similarly  for  the  other  functions.  Or,  otherwise,  the 
triangle  of  reference  for  (<£  —  180°)  is  identical  with  the 
triangle  of  reference  for  (</>  -f-  180°),  and  the  functions  of 
(<£  —  180°)  are  the  same  as  the  functions  of  (<f>  +  180°). 

§  35.  If  360°  is  added  to  <£,  the  triangles  of  reference  for 
<£,  and  (<£  +  360°)  coincide,  and  the  functions  of  <£,  and 
(<£  +  360°)  are  identical,  and  we  have  : 


sin  (360°  —  ^ 

•)  =  sin  (— 

.'.esc  (360°  —  4 

.)  =  —  esc  < 

cos  (360°  —  <£ 

,)  =  cos  (— 

.'.sec  (360°  —  4 

.)  =  sec  <£. 

tan  (360°  —  <£ 

•)  =  tan  (  — 

.\ctn  (360°  —  <£ 

•)  =  —  ctn  < 

28  PLANE  TRIGONOMETRY. 

sin  (360°  +  4>)  =  sin  <£,  .'.  esc  (360°  +  <£)  =  esc  <£. 
cos  (360°  +  <£)  =  cos  <£,  .-.  sec  (360°  +  <£)  =  sec  <£. 
tan  (360°  -f  0)  =  tan<£,  .'.  ctn  (360°  +  <£)  =  ctn  <£. 

Keplacing  <£  by  (—  <£)  these  formulas  give  by  §  32, 

=  —  sin  <£, 

=  cos  <£, 
=  —  tan<£, 

Observing  that  (<£  —  360°)  =  —  (360°  —  <£),  we  have  by  the 
formulas  of  §  32,  and  the  present  section  : 

sin  (<£  —  360°)  =  —  sin  (360°  —  <£)  =  sin  <j>. 

Similarly  for  the  other  functions. 

The  functions  of  (<^>  —  360°)  can  also  be  directly  seen  to 
be  identical  with  the  functions  of  <£,  since  the  triangles  of 
reference  of  the  two  angles  coincide. 

§  36.  We  will  now  consider  the  cases  of  (m  90°  =b  <£)  where 
m  =  1,  and  3.  If  90°  is  added  to  the  angle  <£  of  any  quadrant, 
the  triangles  of  reference  for  (<£  +  90°)  and  <£,  are  geometri- 
cally equal.  The  hypotenuses  are  equal,  the  perpendicular 
and  the  base  of  the  triangle  of  reference  for  (<£  +  90°)  are 
equal  respectively  to  the  base  and  the  negative  of  the  perpen- 
dicular of  the  triangle  of  reference  for  <£.  It  follows  that  any 
function  of  (<£  +  90°)  is  the  co-function  of  <£,  and  that  it  has 
the  opposite  or  the  same  sign  according  as  the  base  of  the 
triangle  of  reference  for  (<£  +  90°)  does,  or  does  not,  enter 
into  the  ratio.  The  following  figure  illustrates  the  preceding 
results. 

We  have  here,       rx  =  r,  y^  =  x,  Xi  =  —  y. 


INTEGRAL  MULTIPLES. 


29 


cos 

tan 


ctn(<£+90°)  =  - 
Y 


•  FIG.  17. 

It  is  to  be  observed  that  the  triangle  of  reference  for 
(<£  +  90)  may  be  obtained  by  revolving  the  triangle  of 
reference  for  <£  about  the  origin  in  the  plane  of  the  figure, 
through  a  positive  angle  of  90°.  This  converts  the  base  into 
the  perpendicular,  and  the  perpendicular  becomes  the  base 
changed  in  sign. 

§37.  If  in  the  formulas  of  §  36,  <£  is  replaced  by  (—  <£), 
they  become  by  the  use  of  the  formulas  of  §  32 : 


sin  (90°  —  <£)  =  cos  (—  <£)  =  cos  <£, 
cos  (90°  —  <£)  =  —  sin  (—  <£)  =  sin  <£, 
tan  (90°  —  <#>)  =  —  ctn  (—  ^>)  =  ctn  <£, 


.'.  esc  (90°  —  <f>)  =  sec  <f>. 
.'.  sec  (90°  —  <f>)  =csc  <#>. 
.-.  ctn  (90°  —  <£)  =  tan  <#>. 


These  formulas  make  clear  the  meaning  of  the  prefix  "  co." 
They  show,  for  example,  that  the  cosine  of  an  angle  <f>  is  the 
complement's  sine,  [sin  (90°  —  <£)].  That  is,  the  word  "  cosine  " 
is  an  abbreviation  for  "  complement's  sine."  Similarly  for 
the  other  functions. 


30 


PLANE   TRIGONOMETRY. 


Since  (<£  —  90°)  =  —  (90°  —  <£),  we  have,  by  the  formulas 
of  §  32,  sin  (<£  —  90°)  =  —  sin  (90°  —  <£)=  —  cos  <£, 

.'.  esc  (<£  —  90°)  =  —  sec  <f>. 

§  38.  If  270°  is  added  to  the  angle  <£  of  any  quadrant,  the 
triangles  of  reference  for  (<£  -f  270°)  and  <£  are  geometrically 
equal,  the  hypotenuses  are  equal,  the  base  and  the  perpen- 
dicular of  the  triangle  of  reference  for  (<£  +  270°)  are  equal, 
respectively,  to  the  perpendicular  and  the  negative  of  the 
base  of  the  triangle  of  reference  for  <£.  It  follows  that  any 
function  of  (<£  +  270°)  is  the  co-function  of  <j>,  and  that  it  has 
the  same,  or  opposite  sign,  according  as  the  perpendicular  of 
the  triangle  of  reference  for  (<£  -j-  270°)  does  not,  or  does, 
enter  into  the  ratio. 

The  following  figure  illustrates  the  preceding  results. 


rl  •=•  r,  xx  =  y,  yt  = 


We  have  here, 

sin  (<£  +  270°)  =      =  — 
ri         r 

cos  O  +  270°)  =  -  =    ^    = 
ri         r 


—  x. 


=  —  cos 


INTEGRAL  MULTIPLES.  31 

tan(<£  +  270°)  =  ^  =  —  =  -  ctn  <£. 

xi      y 

.'.  esc  (<f>  +  270°)  =  —  sec  <£,         sec  (<£  -f  270°)  =  esc  <£, 
ctn  (<£  +  270°)  =  —  tan  <£. 

§  39.  The  angle  (360°  -f  </>)  may  be  obtained  by  adding 
180°  to  the  angle  (<f>  -f 180°)  of  §  33,  whence  the  triangle  of 
reference  for  (<£  -f- 180°)  is  turned  as  before,  through  a  positive 
angle  of  180°,  and,  therefore,  its  base  and  perpendicular  suffer 
a  second  reversal.  And,  in  general,  the  addition  of  m  180°  to 
0,  produces,  or  does  not  produce,  a  reversal  of  the  signs  of  the 
perpendicular  and  base  of  the  triangle  of  reference,  according 
as  m  is  an  odd  or  even  integer,  positive  or  negative. 

The  angle  (<£  +  270°)  may  be  obtained  by  adding  180°  to 
(<£  +  90°)  of  §  36.  This  turns  the  triangle  of  reference,  as 
before,  through  180°,  and  reverses  the  signs  of  both  the  base 
and  perpendicular  of  the  triangle  of  reference  for  (<£  -}-  90°). 
The  perpendicular  and  base  of  the  triangle  of  reference  for  <j> 
have  been  converted  respectively  into  the  base  and  perpen- 
dicular of  the  triangle  of  reference  for  (<£  +  270°),  and  have 
suffered,  respectively,  two,  and  one,  reversals  of  sign.  And,  in 
general,  the  addition  of  (2  n  +  1)  90°  to  <j>,  converts  the  base 
and  perpendicular  of  the  triangle  of  reference  for  <£,  into 
the  perpendicular  and  base,  respectively,  of  the  triangle  of 
reference  f  or  <£  +  (2  n  +  1)  90°.  And  the  base  of  the  result- 
ing triangle  of  reference  has,  or  has  not,  while  the  perpen- 
dicular has  not,  or  has,  been  reversed  in  sign,  according  as  n 
is  even  or  odd. 

§  4O.  It  should  be  carefully  observed  that  in  the  preceding 
results  the  functions  of  any  angle  combined  with  a  multiple 
of  90°  are  equal  numerically  to  the  same,  or  to  the  co-functions 
of  the  angle,  according  as  the  multiplier  is  one  of  the  even 
numbers,  0,  2,  4 ;  or  one  of  the  odd  numbers,  1,  3. 

It  can  be  seen  from  the  discussion  of  the  next  section  that 
the  following  general  relations  exist : 


32  PLANE   TRIGONOMETRY. 

(        TT          \ 
Function  (  2  n  —  ±  <#>  )  =  Function  <£,  numerically, 

Function  (  (2  n  -j-  1)  —  ±  <£  J  =  Co-function  <£,  numerically, 

where  n  is  zero,  or  any  integer,  positive  or  negative. 

§41.    We  will  now  give  a  general  demonstration  of  the 

relations  of  the  functions  of  (  m  —  ±  </>  )  to  the  functions  of  <£, 

\     J         / 

where  m  is  any  integer,  positive  or  negative.      Let  x^  yi, 
r1?  and  x,  y,  r,  respectively,  be   the   sides  of  the   triangles 

7T      '"• 

of  reference  for  <£  +  m  —  >  and  <#>.     The  problem  divides  itself 
& 

into  two  cases : 

a.  When  m   is  Even,  and  Equal  to  2  n. 
We  have  by  §  39  : 


r  =  tan  <£, 

2V       *i       (-l)nx 


n  77  +  <A   1  =  Ctn  ^>- 


INTEGRAL  MULTIPLES.  38 

§  42.   Replacing  <f>  by  (—  <£)  in  the  formulas  of  §  41,  we  get 
by  the  formulas  of  §  32  : 


sin     2  n  —  —  *     =  (—  1)    sm  (—  *)  =  —  (—  1)    sm 


.  .  esc 


cos  I  2  n  -  —  *  1  =  (—  1)   cos  (—  *)  =  (—  1)  '  cos 
\  / 

.'.  sec  f2n£-*  )  =  (-!)"  sec*. 


tan  f  2  n  -  —  <£  )  =  tan  (—  <f>)  =  —  tan 


.'.  ctn     2  n  -  —  e£     =  —  ctn  <£. 
A 


Formulas  for  functions  off*  —  2n-J  are  covered  by  the 
formulas  of  §  41,  since  n  may  be  negative  as  well  as  positive. 

(  ir\ 

Or,  otherwise,  the  triangle  of  reference  for  (  *  —  2  n  —  )  is 

identical  with  the  triangle  of  reference  f or  (  *  +  2  n  —  )  • 

\  *  / 

b.  When  m  is  Odd,  and  Equal  to  (2  n  +  1). 
§  43.   We  have  by  §  39  : 


r  =  r. 


ri 
csc  (  (2  n  +  1)  £  +  *)  =  (-  l)n  sec 


34  PLANE  TRIGONOMETRY. 


.-.  sec  ((2  n 


+  1)      +        =    -  (-  1)"  esc  </>. 


.'.  ctii  (  (2  n  +  1)  |  +  <£  J  =  -  tan  <£. 

§  44.    Replacing  <£  by  (  —  <£)  in  the  formulas  of  §  43,  we 
have  by  the  formulas  of  §  32  : 


sin     (2  n  +  1)      -        =  (-  l)n  cos  (-  *)  =  (-  l)n  cos  <#>, 

esc  ^(2  n  +  1)  |  -  ^  -  (-  l)n  sec  <j>. 

cos     (2  n  +  1)      -        =  -  (-  l)n  sin  (-  <£)  -  (-  l)n  sin  <^> 


/.  sec         n 


tan  f  (2  n  +  1)  ^  —  <^>  J  =—  ctn  (-  </>)  =  ctn  <fc 


Formulas  for  functions  of  (  <#>  —  (2  n  -\-  1)  —  1  are  covered 

by  the  formulas  of  §  43,  since  n  may  be  negative  as  well  as 
positive. 

§  45.    The  results  of  §§  31-44  inclusive,  are  contained  in 
the  following  formulas  : 


INTEGRAL  MULTIPLES.  35 


Sin      z  n  — 


cos     2  n      ±  <£     =  (—  l)n  cos  <£, 

tan  (  2  n  ^  ±  <£  )  =  ±  tan  <£, 
\       J         / 

/  TT         \ 

sin  f  (2  n  +  1)  -  ±  <H  =  (-l)n  cos  <#>, 

cos  f  (2  n  +  1)  |  ±  </>")  -  =p  (-  l)n  sin  </>, 
tan  ^  (2  n  +  1)  ^  ±  <#>  j  =  +  ctn  <£. 

§  46.   The  formulas  of  §  45  may  be  put  into  the  following 
form  : 

Since  (—  l)n  sin  <f>  =  sin  [(—  l)n  <£]  whether  n  is  even  or  odd, 
(—  l)n  cos  <£  =  cos  [(—  l)n  <£]  only  when  n  is  even, 
(—  l)n  tan  <£  =  tan  [(—  l)n  </>]  whether  n  is  even  or  odd, 

we  have,  dividing  by  the  sign  factor  of  the  second  member  of 
the  first  three  formulas  of  §  45,  and  writing  the  second 
members  first  : 


sn 


=  ±  (—  l)n  sin  f  2  n  ^ 


cos  <j>  =  (—  l)n  cos     2  n      ± 


=  cos  (  4  m  —  ±  <^>  )  where  2  m  =  n, 
\        z 


36  PLANE   TRIGONOMETRY. 


—  tan  (  ±  2  n  ^ 


tan  <f>  =  ±  tan     2n±<£    —  tan     ±  2  n      +  <f> 


Since  n  may  be  positive  or  negative  at  pleasure,  the 
preceding  can  be  written  more  simply  : 

sin  <£  =  sin  (n  TT  +  (—  l)n  </>), 
cos  <£  =  cos  (2  m  TT  ±  <£), 
tan  <£  =  tan  (n  TT  -j-  <£). 

§47.  If  in  the  formulas  of  §  46,  (n7r+(—  1)»,  (2m7r±<£), 
(n7r-]-<£),  be  successively  denoted  by  3>,  then  $  represents  all 
possible  angles  whose  sines,  cosines,  and  tangents,  are  respec- 
tively equal  to  the  sine,  cosine,  and  tangent  of  the  particular 
angle  <£.  Thus,  if  sin  <£  =  -J-,  the  simplest  angle  may  be 
constructed  whose  sine  is  -J-,  and  the  general  solution  is 
$  =  n  TT  +  (—  l)n  <£>  where  <£  is  the  angle  constructed. 


B.  Periodicity  of  the  Trigonometric  Functions. 

§  48.  When  f  (x  +  na}=  f  (x),  where  n  is  zero,  or  any  posi- 
tive or  negative  integer,  f  (x)  is  said  to  be  periodic,  and  "  a  " 
is  called  the  period  of  f  (x).  From  the  formulas  of  §  41,  it 
appears  that  if,  in  them,  n  =  2  m,  where  m  is  zero,  or  any 
positive  or  negative  integer, 

sin  (<£  -f-  2  m  TT)  =  sin  <£,         esc  (<£  +  2  m  TT)  =  esc  fa 
cos  (<j>  +  2  m  TT)  =  cos  fa         sec  (<£  +  2  m  TT)  =  sec  fa 

and  that  always, 

tan  (^  -f-  n  TT)  =  tan  <£,  ctn  (<£  -f-  n  TT)  =  ctn  fa 

and  that,  therefore,  the  period  of  the  sine,  cosine,  secant,  and 
cosecant  is  2  TT  ;  and  of  the  tangent  and  cotangent  is  TT. 


> 


THE  FUNCTIONS   OF  SOME  SPECIAL  ANGLES. 


37 


r.   The  Functions  of  Some  Special  Angles. 

§  49.  To  find  the  functions 
of  30°.  Let  XOP  be  equal  to 
30°,  and  let  OMP  be  its  tri- 
angle of  reference. 

Produce   PM  to   P',  making 
MP'  =    PM,   and   draw    OP'. 
The  triangle  OPP'  is  geomet- 
p/  rically    equilateral,    and     MP 

FIG.  19.  =iOP. 


OM  =  VOP2-MP2  = 
From  this  we  get, 


==^L=I.^L=:±,    .-.esc  30°  =  2. 


./q 


Vs 


sec  30°  =  f  V3. 
ctn30°  =  V5. 


§  5O.  To  find  the  func- 
tions of  45°.  Let  XOP  be 
equal  to  45°,  and  let  OMP 
be  its  triangle  of  reference. 
By  geometry  OM  =  MP,  and 
OM2  +  ¥P2  =  2  OM2  =  2  M~P2 


= 

V2 


f 

'•/ 

HP 

Y 

0 

M 

Y' 

FIG.  20. 

38  PLANE   TRIGONOMETRY. 


From  this  we  get, 

-Up 

.     ,eo      MP      V2  1        1    /-  ,eo        /- 

Sm45   ;==  =        =     V2    •••CSC45°  = 


-UP        - 

_0      OM      V2  11    /-  ._        /- 

cos  46"==        =  --  =       =  ;V2,  ,.  sec  45°  = 


tan  45°  =         =  1,  /.  ctn  45°  =  1. 


Or  otherwise  thus  :  Since  45°  —  90°  —  45°,  we  have, 
sin  45°  =  sin  (90°  —  45°)  =  cos  45°. 
sin2  45°  +  cos2  45°  =  2  sin2  45°  =  2  cos2  45°  =  1, 

.'.  sin  45°  =  cos  45°  =  4-  =  ^  V2. 

V2      2 

The  other  functions  of  45°  may  be  obtained  from  the  sine 
and  cosine  of  45°.  Since  the  angles  60°,  120°,  150°,  210°, 
240°,  300°,  330°,  are  cases  of  (m  90°  ±30°);  and  the  angles 
135°,  225°,  315°,  are  cases  of  (m  90°  ±  45°),  therefore  their 
functions  can  be  at  once  obtained  in  terms  of  the  known 
functions  of  30°  and  45°,  respectively,  by  the  use  of  the 
formulas  of  §§  33-38,  inclusive. 


D.  Examples.     IV. 

1.  Find  the  complements  and  the  supplements  of  the  following  angles: 
57°,  -  210°,  -  65°,  -  116°,  3r,  -  2M5. 

Find  the  functions  of  the  following  angles  : 

2.  60°;      3.   120°;      4.   135°;      5.   150°;      6.  210°;     7.  225°;     8.  240°; 
9.  300°;     10.  315°;     11.  330°. 

For  convenience  of  reference  the  following  table  is  given  : 


EXAMPLES.    IV. 


39 


Angles. 

Functions. 

sin 

cos 

tan 

ctn 

sec 

CSC 

0°,  360° 

180° 

0 

±1 

0 

QO 

±1 

CO 

30°,  150° 
210°,  330° 

(±H 

(+-HV3 

(+-HV3 

(+-)V3 

(+-)*V3 

(±)2 

45°,  135° 
225°,  315° 

(±)iVi 

(+-HV2 

(+-)! 

(-f-)l 

(+_)V2 

(±)V2 

60°,  120° 
240°,  300° 

(±)iV3 

(+-H 

(+-)V3 

(+-)*V3 

(+_)2 

(±)fV3 

90° 
270° 

±1 

0 

00 

0 

00 

±1 

In  this  table  (±)  is  an  abbreviation  for  (±  ±),  (+_)  for  (±  =j=),  and 


12.  Find  the  functions  of  the  following  angles :  —  30°,  420°,  765°,  510°, 
-  585°,  3360°,  1260°. 

Prove  that, 

13.  sin  295°  =  cos  155°. 
tan  217°  =  ctn  (—  127°). 
sec  (-  37°)  =  -  esc  233°. 
cos  11 45°  =  sin  (-205°). 
ctn  (-  127°)  =  -  tan  503°. 
esc  (—  464°)  =  sec  1606°. 

Given  tan  200°  =  0.364,  find  sec  70°. 
1 


14. 

15. 
16. 
17. 

18. 
19. 


20.  Given  sin  165°  =  — -=.  ( VI  —  1),  find  tan  285°. 

2V2 

If  A,  B,  and  C  are  the  angles  of  a  plane  triangle,  prove  that : 

21.  sin  A  =  sin  (B+C),  cos  A  =  —  cos(B  +  C),  tan  A  =  —tan  (B+C), 
sin  A/2  =  cos  J  (B  +  C),  cos  A/2  =  sin \ (B  +  C),  tan  A/2  =  ctn |(B  +  C). 

22.  sin  A/2  sin  B/2  sin  C/2  =  cos  \  (B  +  C)  cos  $  (C  +  A)  cos  i  (A+  B). 

23.  cos  A/2  cos  B/2  cos  C/2  =  sin -j-  (B  +  C)  sin  i  (C  +  A)  sin  i  (A  +  B). 

24.  tan  A/2  tan  B/2  tan  C/2  =  ctn  -J-  (B  +  C)  ctn  -J-  (C  +  A)  ctn  •£  (A+  B). 

25.  sin  (A  +  B  +  C)  =  0,  cos  (A  +  B  +  C)  =  -  1,  tan  (A  +  B  +  C)  =  0. 

26.  sini(A+B  +  C)  =  l,  cos|(A+  B  +  C)  =  0,  tan£(A+  B+  C)  =  oo  . 

27.  sin  A  =  -  sin  (2  A  +  B  +  C). 


40  PLANE   TRIGONOMETRY. 

28.  sin  |  (A  -  B)  =  cos  i  (2  B  +  C),  tan  i  (3  A  +  B)  =  ctn  |  (C  -  2  A). 

29.  sin  n  (A  +  B)  -  —  (—  l)n  sin  nC. 

30.  cosn(A'+  B)=  (-l)ncosnC. 

31.  tan  n  (A  +  B)  =  —  tan  nC. 

Find  an  expression  for  all  possible  angles : 

32.  Whose  secants  are  the  same  as  sec  0. 

33.  Whose  cosecants  are  the  same  as  esc  0. 

34.  Whose  cotangents  are  the  same  as  ctn  0. 

35.  Obtain  formulas  for  the  functions  of  (0  —  2  n  rt/2)  in  terms  of 
the  functions  of  0,  where  n  is  always  a  positive  integer. 

36.  The  same  for  [0  —  (2  n  +  1)  7t/2~\. 

37.  Obtain  the  functions  of  2  n  jr/2,  (2  n  + 1)  it/2,  4  n  if/2,  (4  n  + 1)  if/2, 
(4  n  -f  2)  if/2,  (4  n  +  3)  if/2. 

NOTE  ON  THE  GENERAL  SOLUTION  OF  TRIGONOMETRIC  EQUATIONS.  — 
A  general  solution  of  a  trigonometric  equation  is  all  possible  angles  which 
satisfy  the  given  equation.  When  the  equation  is  algebraic  in  form,  the 
general  solution  may  be  conveniently  expressed  by  a  finite  number  of 
general  formulas,  each  of  which  gives  an  infinite  number  of  values. 
Thus  if  the  given  equation  be  sin  0  =  -£,  an  expression  containing  all 
possible  angles,  whose  sines  are  i,  is  n?r  +  (—  l)n  if/Q  (by  formulas  of 
§47),  and  «t>  =  mt  +  (—  l)n  ?r/6  is  the  general  solution.  This  is  an 
algebraic  equation  of  the  first  degree  in  terms  of  sin  0,  and  there  is  one 
general  formula  which  contains  an  infinite  number  of  values  in  itself. 

If  2  sin20  —  3  sin  0  +  1  =  0  is  the  given  equation,  we  have  : 

(2  sin  0  —  1)  (sin  0  —  1)  =  0,        sin  0  =  •£,        sin  0  =  1, 
*i=  nx+  (-l)n7T/6,  $2  =  r\Tt  +  (-  l)n  if/2, 

as  the  general  solution.     And  it  will  be  found  that  the  number  of  general 
formulas  is  not  greater  than  the  degree  of  the  equation. 
Obtain  general  solutions  of  the  following  equations  : 

38.  sin  0  =  Vj. 

39.  cos  0  =  3-. 

40.  tan  0  =  —  Vfr. 

41.  sec20  —  tan  0  —  1  =  0. 

42.  2  sin  0  cos  0  —  2  sin  0  +  cos  0  —  1  =  0. 

43.  2  sin  0  =  esc  0. 

44.  3  tan  0  —  ctn  0  =  0. 

45.  4  cos  0  =  sec  0. 

46.  3  sin20  +  cos>20  —  sin  0  =  0. 

47.  cos30  —  6  cos20  +  1 1  cos  0  —  6  =  0. 


EXAMPLES.    IV.  41 

Show  that  the  following  general  formulas  give  the  same  sets  of  values, 
though  not  corresponding  value  by  value,  for  the  same  value  of  n. 

48.  $  =  n?r  +  (—  l)n  5  x/Q,  and  2  ntf  +  7t/2  ±  it/?,. 

49.  $  =  -  7T/4  -  nx  +  (-  l)n  tf/4,  4>  =  7T/4  +  2  ntf  ±  Tt/±. 

50.  Show  that  one  value  of  each  of  the  expressions,  sin-1  x  +  cos-1  x, 
tan-1  x  +  ctn-1  x,  sec-1  x  +  esc-1  x,  is  jf/2. 

51.  Find  one  value  of  each  of  the  following  expressions:  sin— 1 3/5  + 

sin-1 4/5,  sec-1 5  +  sec-1  -^-= ,  tan-1 3  +  tan-1 1/3. 

2V6 

52.  If  0  is  the  simplest  value  of  sin—1  x,  sec~ l  x,  tan—1  x,  successively, 
prove  that  in  general, 

sin-1  x  +  cos-1  x  =  [n;r  +  (—  l)n  0]  +  [2  tntf  ±  (a/2  —  0)] 
=  [2  (2  m  +  n)  ±  1]  |  +  [(-  l)n  ±  1]  0. 

sec-i  x  +  csc-ix  =  [  2  mt  ±  0]  4-  [nit  +  (-  l)m  (?r/2  -  0)] 

=  [2  (2  n  +  m)  +  (-  l)mj  7t/2  +  [(-  l)m+1±l]  0. 

tan-1  x  +  ctn-1  x  =  [mt  +  0]  +  [IUTT  +  7t/2  —  0] 
=  [2  (m  +  n)  +  1]  7T/2, 

where  m  and  n  may  be  0,  or  any  positive  or  negative  integers,  and  are 
independent  of  each  other.  How  may  m  and  n  and  the  double  sign  be 
taken  to  give  the  particular  solutions  of  problem  50  ? 


CHAPTER   V. 

A.    Projection. 

§  51.  The  orthogonal  projection  of  a  finite  line  upon  an 
axis  of  reference  is  the  segment  of  the  axis  taken  from  the 
foot  of  the  perpendicular  from  the  origin  of  the  line  upon  the 
axis,  to  that  from  the  term. 

It  will  be  proved  that  such  projection  of  a  straight  line  is 
equal  to  the  length  of  the  line  multiplied  by  the  cosine  of  the 
angle  between  the  positive  directions  of  the  axis  and  the  line. 
For  convenience,  projection  will  be  considered  under  two 
cases : 

a.  When  the  Line  and  Axis  are  Co-planar. 

Thus,  when  the  line  OP  and  axes,  OX  and  OY,  lie  in  the 
same  plane,  the  projections  of  OP  on  the  axes,  or  fixed 
parallel  lines,  are  the  segments  of  the  axes,  or  fixed  parallel 
lines,  taken  from  the  feet  of  the  perpendiculars  dropped  upon 
them  from  0,  to  those  from  P.  These  segments  are  obviously 
the  base  and  the  perpendicular,  respectively,  of  the  triangle 
of  reference  for  the  angle  XOP. 


FIG.  22. 


PROJECTION. 


43 


\ 


In  Figs.  21-24  we  have 

YOP  -  YOX  +  XOP  =  -  90°  +  $  +  n  360°. 

MP      v 
sin  <£>  —  -^p-  =  L ;  sin  <j>  =  cos  (<j>  —  90°  -f  n  360°). 

OM      x 
cos  <£  =  -^_-  =  -  •       And  these  give 

y  =  r  cos  (<£  —  90°  +  n  360°),  and  x  =  r  cos  <£, 
or  MP  =  OP  cos  YOP,  and  OM  =  OP  cos  XOP. 

Since  n  may  be  0  or  any  integer  without  affecting  the 
preceding  result,  it  will  be  so  chosen,  for  uniformity,  that 
the  angle  YOP  shall  be  always  the  simplest  positive  angle. 

b.  When  the  Line  and  Axis  are  not  Co-planar. 

§  52.  When  the  line  and  axis  are  not  co-planar,  the 
angle  between  them  is  the  angle  taken  from  the  positive 
direction  of  the  axis  to  the  positive  direction  of  a  line 
parallel  to  the  given  line,  and  co-planar  with  the  axis.  It 
is  evident  that  if  two  planes  o  and  p  be  passed  respectively 
through  the  extremities  0  and  P  of  the  line  OP,  and  perpen- 
dicular to  the  axis,  the  segment  of  the  axis  from  plane  o 
to  plane  p,  is  the  projection  of  OP  on  the  axis. 

In  the  figure,  O'P'  is  parallel  to  OP,  cuts  the  axis,  and 
pierces  o  and  p  in  0'  and  P',  respectively.  O'M  and  P'M" 
lie  in  o  and  p,  and  cut  the  axis  at  M  and  M'. 


44 


PLANE  TRIGONOMETRY. 


0' 


FIG.  25. 


By  geometry,  0'P'  =  OP,  0'M"=  MM'.     O'M  and  P'M"  are 
perpendicular  to  MX. 
'   By  a.  0'M"  =  0'P'cos<£,  or  MM'  =  OPcos<£. 

§  53.  It  follows  from  the  preceding  work  that  the  projec- 
tion of  the  straight  line  0  P  upon  any  axis  is  equal  to  the  sum 
of  the  projections  of  the  parts  of  an  uninterrupted  broken 
line  going  from  0  to  P. 

The  following  notation  will  be  used.  The  projection  of  a 
line,  I,  on  an  axis,  a,  will  be  denoted  by  la. 


B.  Functions  of  Sums  and  Differences  of  Angles. 

§  54.   In  this  figure,  from  the  triangle  of  reference  for  /2, 


FIG.  26. 


FUNCTIONS  OF  SUMS  AND  DIFFERENCES  OF  ANGLES.    45 


cos  (3  =    -    »         .-.  OM  =  OP  cos  ft. 


By  §  53,  OPy  =  OMy+MPy. 


These  become  : 

OP  sin  (a  +  /5)  =  OM  sin  a  -f  M  P  cos  a,* 
or,  by  substituting  the  values  of  OM  and  MP, 

OP  sin  (a  +  0)  =  OP  sin  a  cos  (3  -f  OP  cos  a  sin  /?. 

.'.  sin  (a  +  /?)  =  sin  a  cos  /8  +  cos  a  sin  /?. 
OP  cos  (a  +  P)  =  OM  cos  a  +  M  P  cos  (90°  +  a)* 

which  reduces  to 

cos  (a  +  /?)  =  cos  a  cos  |8  —  sin  a  sin  /?. 

§  55.  If  in  the  formulas  of  §  54  ft  is  replaced  by  (—  /?) 
we  obtain  : 

sin  (a  —  ft)  =  sin  a  cos  /3  —  cos  a  sin  /?. 
cos  (a  —  /8)  =  cos  a  cos  /3  -f-  sin  a  sin  /?. 

§  56.   From  the  formulas  of  §§  54  and  55 

_^_  sin  (a  =b  /3)  _  sin  a  cos  /3  ±  cos  a  sin  /8 

~~    ^      cos  (a  ±  /3)      cos  a  cos  /S  =p  sin  a  sin  y8 

sin  a  cos  /3     cos  a  sin  /? 
_  cos  a  cos  /3  ~~  cos  a  cos  /3  _  tan  a  ±  tan  ^8 

cos  a  cos  /3      sin  a  sin  (3      1  =p  tan  a  tan  fi 

cos  a  cos  /8  ^~  cos  a  cos  /? 

*  The  values  of  these  projections  are  in  all  cases  justified  by  the 
principle  that  if  a  line  taken  in  the  opposite  direction,  and  for  the  sake 
of  the  triangle  of  reference  for  /3,  has  its  sign  changed,  then  the  projec- 
tion becomes  —  I  cos  (0  ip  180°)  =  I  cos  0,  generally  as  before,  with  the  I 
essentially  negative. 


46  PLANE   TRIGONOMETRY. 

/    _4_  o\  —          1          _  1  =F  tan  a  tan  (3 
™     tan  (a  ±  /?)  ~  tan  a  ±  tan  j3 
_  ctn  a  ctn  /?  =F  1 
ctn  /3  ±  ctn  a 

C.   Functions  of  Multiple  Angles. 

§  57.   If  in  the  formulas  of  §§54  and  56  ft  is  replaced  by  a, 

sin  (a  -f-  a)  =  sin  2  a  =  sin  a  cos  a  +  cos  a  sin  a  =  2  sin  a  cos  a. 
cos  (a  -f"  a)  =  cos  2  a  =  cos  a  cos  a  —  sin  a  sin  a  =  cos2a  —  sin2a 
—  2  cos2a  —  1  =  1  —  2  sin2a. 

tan  a  +  tan  a         2  tan  a 

tan  ( a  -+•  a }  =  tan  Z  a  = : = r-  • 

1  —  tan  a  tan  a      1  —  tan2a 

ctn  a  ctn  a  —  1      ctn2a  —  1 


ctn  (a  +  a)  =  ctn  2  a  = 


ctn  a  +  ctn  a          2  ctn  # 


/>.   Functions  of  Sub-Multiple  Angles. 

§  58.   By  the  formulas  of  §  57 

cos  2  a'  =  1  —  2  sin  V  =  2  cosV  - 1. 
If  in  these  formulas  2  a'  is  replaced  by  a, 

cos  a  =  1  —  2  sin2  - >    .'.2  sin2  -  =  1  —  cos  a. 


1  —  cos  a 


And         cos  a  =  2  cos2 -  —  1,     .'.  2  cos2-  =  1  -|-cosa. 

—  w 


a^Jl  +  cosa. 
2 

1  —  cos  a 

1  +  cos  a 


cos  a 


—  cos  a 


FORMULAS.  47 

E.   Formulas  for  the  Sums  and  Differences  of  the 

Sines  of  two  Angles,  also  of  the  Cosines 

of  two  Angles. 

§  59.   By  §§  54  and  55 

sin  (a'  +  ft')  +  sin  (a'  —  ft')  =2  sin  a'  cos  £'. 
sin  (a'  +  ft')  —  sin  (a1  —  ft1)  =  2  cos  a'  sin  /?'. 
cos  (a1  -f  )3')  +  cos  (a'  —  ft')  =  2  cos  a'  cos  0'. 
cos  (a'  —  £')  —  cos  (a1  +  ft')  =  2  sin  a'  sin  ft'. 

In  these  formulas  let     a'  +  /?'  =  a, 


and  the  formulas  become  : 


sin  a  +  sin  /?  =  2  sin  -J  (a  -f-  /3)  cos  %  (a  —  ft). 
sin  a  —  sin  /5  —  2  cos  -J-  (a  +  ft)  sin  -J-  (a  —  /8). 
cos  a  -h  cos  ^8  =  2  cos  |(a  +  ft)  cos  £  (a  —  ft). 
cos  /3  —  cos  a  =  2  sin  £  (a  -f-  )8)  sin  i  (a  —  ft). 

sin  a  +  sin  ^  _  tan  £  (a 


sin  a  —  sin  ^8      tan|-(a  — 

sin  a  +  sin  ft 

—  = 
cos  a  +  cos  ^8 

sin  a  +  sin 


cos  ft  —  cos  a 
sin  a  —  sin  ft 
cos  a  +  cos  ft 
sin  a  —  sin  ft 
cos  ft  —  cos  a 
cos  a  -\-  cos  ft 
cos  8  —  cos  a 


=  ctn  -J-  (a  —  j8). 

N 
=  tan  ^(a  —  ft). 


=  ctn  ^  (a  +  0)  ctn  %(a  — 


48  PLANE   TRIGONOMETRY. 


F.   Examples.     V. 

Find  the  functions  of  : 

1.    15°.  2.    22|°.  3.    37£°.  4.    75°. 

Prove  : 

5.  ctn  2  a  =•  %  (ctn  a  —  tan  a). 

6.  esc  2  a  =  %  (tan  a+  ctn  a). 

7.  tan  a  =  esc  2  a  —  ctn  2  or. 

8.  ctn  a  =  esc  2  a  +  ctn  2  or. 

9.  tan  (45°  +  $  a)  =  sec  a  +  tan  <*  =  ctn  (45°  —  -£-  a). 

10.  tan  (45°  -  i  a)  =  sec  a  -  tan  a  =  ctn  (45°  +  i  a). 

11.  tan  (45°  +  i  a)  tan  (45°  -  i  a)  =  1. 

12.  tan  50°  +  ctn  50°  =  2  sec  10°. 

13.  tan  50°  —  ctn  50°  =  2  tan  10°. 

14.  cos  (60°  +  a)  +  cos  (60°  —  a)  =  cos  a. 

15.  sin  (30°  +  or)  +  sin  (30°  —  a)  =  cos  a. 

16.  sin  (45°  +  a)  =  cos  (45°  -  a). 
Find: 

17.  sin  (a  +  0  +  7). 

18.  cos  (a +  0  +  7). 

19.  tan(a  +  p  +  y). 

20.  sin  3  a  in  terms  of  sin  a. 

21.  cos  3  a  in  terms  of  cos  or. 

22.  tan  3  a  in  terms  of  tan  a. 

23.  sin  4  a:  in  terms  of  sin  a  and  cos  a. 

24.  cos  4  a:  in  terms  of  cos  a. 

25.  tan  4  <*  in  terms  of  tan  a. 

26.  sin  5  a  in  terms  of  sin  a. 

27.  cos  5  a  in  terms  of  cos  a. 
Prove : 

28.  sin  a  +  sin  /3  +  sin  7  —  sin  (a  +  0  +  7) 

=  4sin£(/3  +  7)sin|(7  +  a)sini(a  + /S). 

29.  cos  a  +  cos  /3  +  cos  7  +  cos  (a  +  /3  +  7) 

=  4  cos  |  (j8  -f  7)  cos  £  (7  +  a)  cos  £  (a  +  /3). 

30.  tan  a  +  tan  p  +  tan  7  —  tan  a  tan  /3  tan  7  =  sm  v* — £ — Zl  • 

cos  or  cos  /3  cos  7 

31.  ctn  a  +  ctn  fl  +  ctn  7  —  ctn  a  ctn  fl  ctn  7  =  —     — —. — -~ -' 

sin  a  sin  p  sin  7 

32.  sin2x  +  sin2y  +  sin2z  +  sin2  (x  +  y  +  z) 

=  2  [1  —  cos  (y  +  z)  cos  (z  +  x)  cos  (x  +  y)]. 


EXAMPLES.     VI.  49 

33.    cos2x  +  cos2y  +  cos2z  +  cos2  (x  +  y  +  z) 

=  2  [1  +  cos  (y  +  z)  cos  (z  +  x)  cos  (x  +  y)]. 


G.   Examples.    VI. 

1.  If  7g,  5°,  and  pr  be  the  three  measures  of  an  angle,  prove  that 

r-,=*>*. 

7t 

2.  A  bicycle  is  propelled  on  a  circular  arc  of  £  mile  radius  at  the  rate 
of  30  miles  per  hour.     Through  what  angle  has  it  turned  in  two  minutes  ? 

Prove  the  following  identities  : 

3. 


csc20      sec20 

4.  sin20  (1  +  n  ctn20)  +  cos20  (1  +  n  tan20) 

=  sin20  (n  +  ctn20)  +  cos20  (n  +  tan20). 

5.  sin60  —  cos60  =  (2  sin20  —  1)  (1  —  sin  0  cos  0)  (1  +  sin  0  cos  0). 
(tan  0-1)  (esc  0  -  1)  ctn  0  +  1  _ 

ctn  0  —  1  (tan  0  +  1)  (esc  0+1) 

7.  Find  sin  18°  and  cos  18°.     (Use  the  geometric  properties  of  the 
decagon.) 

8.  Find  sin  54°  and  cos  54°. 
Prove  the  following  identities  : 

9.  sin  65°  +  cos  65°  =  V2  cos  20°. 
10.    sin  78°  —  sin  18°  +  cos  132°=  0. 

sin  0  sin  2  0  +  sin  2  0  sin  5  0  +  sin  3  0  sin  10  0  _ 

cos  0  sin  20  +  sin  2  0  cos  50  —  cos  3  0  sin  10  0 

sin  0  +  sin  3  0  +  sin  5  0  +  sin  7  0 
"    cos  0  +  cos  3  0  +  cos  5  0  +  cos  7  0  ~  *" 

3    sin  0  sin  2  0  +  sin  0  sin  4  0  +  sin  2  0  sin  7  0  _ 

sin  0  cos  2  0  +  sin  2  0  cos  5  0  +  sin  0  cos  8  0  ~~ 

eoe(a-8ffi--co.(8a-fl  _ 

sm  2  «  +  sin  2  (3 


16.  =  2co    g_ 

sm  2  a  +  sm  2  /3 

„     a  sin  (a  —  ft)  +  b  sin  or  +  a  sin  (or  +  ft)  _ 
a  cos  (a  —  /3)  +  b  cos  a  +  a  cos  (a  +  j8)  ~~ 

17.  —  sin  (a  +  ft  +  7)  +  sin  (—  a  +  /3  +  7)  +  sin  (a  —  /S  +  7) 

+  sin  (a  +  j8  —  7)  =  4  sin  a  sin  j3  sin  7. 

18.  cos  (a  +  j8  +  7)  +  cos  (—  a  +  p  +  7)  +  cos  (a  —  0  +  7) 

+  cos(a:  +  £  —  7)  =  4  cos  a  cos  0  cos  7. 


50 


PLANE  TRIGONOMETRY. 


19.    8  cos 


a  + 


y        — 


cos 


—  7 
- 


2  2  2 

=  cos  2  a  4-  cos  2  /3  +  cos  2  7  +  4  cos  a  cos  /3  cos  7  +  1. 

ctn  (n  —  2)  0  ctn  n  0  +  1 

20.  s -A. li_  — r =  £(ctn0  —  tan0). 

ctn  (n  —  2)  0  —  ctn  n  0 


21. 


sin  n  0  —  sin  (n  —  2)  0  _ 


=  ctn  (n  —  1)  0. 


cos  (n  —  2)0  —  cos  n  0 
Justify  the  following  statements  : 


22.  sin-1  a  ±  sin-1  b  =  sin*1  [a  Vl  —  b2  ±  b  Vl  -  a2]. 

23.  cos-1  a  ±  cos-1  b  =  cos-1  [ab  ^  V(l  -  a2)  (1  -  b2)  ]. 

a±b 


24.    tan-1  a  ±  tan-1  b  =  tan-1 


ab 


25.  tan-1  a  +  tan-1  b  +  tan-1  c  =  tan-1 

26.  tan-1  a  +  tan-1  b  +  tan-1 


a 


b  +  c-abc 


1  —  ab —  be  —  ca 
1  — a— b  — ab_g 
4* 


1+a+b- 


27.    a  VI  -  b2  +  b  VI  -  a2  =  1,  given  sin-1  a  +  sin-1  b  =  £ 


28.  sin-1  a  -  sin-1  b  =  cos-1  [ab  +  Vl  -  a2  -  b2  +  a2b2]. 

29.  a*  =  i,  given  tan-1 |±-|  +  tan-1 1^|  =  * 

30.  a  +  b  +  c  =  abc,  given  tan-1  a  +  tan-1  b  +  tan-1  c  =  it. 

31.  2  sin  ^  =  +  Vl  +  sin  0  +  Vl  —  sin  0,  given  0  =  200°. 


32.  2  sin  |  =  —  Vl  +  sin  0  —  Vl  —  sin  0,  given  450°  <  0  <  630°. 

33.  2  cos  |  =  —  Vl  4-  sin  0  +  Vl  —  sin  0  between  fixed  limits.    Deter- 
mine the  limits. 

Solve  for  0  in  the  following  equations : 

34.  a  cos  0  +  b  sin  0  =  1. 

Suggestion :    a  cos  0  +  b  sin  0  =  a  (  cos  0  +  -  sin  0  Y      Put  -  =  tan  a 

and  get  cos  (0  —  or)  =  —  — ,  from  which  the  general  value  of  0  can  be 

found. 

35.  sin  0  +  cos  0  =  1. 

36.  sin  0  — cos  0  =  1. 

37.  sin  0  +  cos  0  =  Vj. 

38.  sin0  +  V3  cos0=  1. 

39.  V3  sin  0  —  cos  0  =  N/2. 


EXAMPLES.     VI.  51 


40.  sin  0  +  2  cos  0  =  -J-. 

41.  cos  (a  +  0)  -  sin  (ex  +  0)  =  V2  cos  /5. 
Construct  the  graphs  of  the  following  expressions : 

42.  sin  0  +  cos  0. 

43.  sin  0  —  cos  0. 

44.  tan0+ctn0. 

45.  tan0— ctn0. 

46.  sec  0  +  esc  0. 

47.  sec  0  —  esc  0. 

48.  sec  0  +  tan  0. 

49.  sec  0  —  tan  0. 


CHAPTER  VI. 


A.  Right  Triangles. 

§61.  To  solve  a  triangle  is  to  find  its  unknown  from  its 
known  parts.  It  is  shown  in  plane  geometry  that  to  deter- 
mine a  triangle  three  parts,  one  of  which  is  a  side,  must  be 
given.  In  right  triangles  the  right  angle  counts  as  one  of  the 
three  given  parts. 

a.  Cases. 

§  62.  The  following  table  gives  all  the  cases  which  can 
arise  in  the  solution  of  right  triangles. 


Case. 

Given. 

Required. 

1 

*,  y 

0,  90°  —  0,  r 

2 

r,  x  or  y 

0,  y  or  x,  90°  -  0 

3 

r.  0  or  90°  —  0 

x,  y,  90°  —  0  or  0 

4 

x  or  y,  0  or  90°  —  0 

r,  90°  —  0  or  0,  y  or  x 

b.   Solutions. 

Since  the  angles  <f>  and  90°  —  <£  are  acute,  and,  therefore,  of 
the  first  quadrant,  all  their  functions  are  positive.     We  will 

Y 


X'- 


M 


Y' 


FIG.  27. 


EXAMPLES,     VII.  53 

express  all  solutions  in  terms  of  the  given  parts,  but  some- 
times, in  order  to  use  logarithmic  computation,  it  will  be 
necessary  to  express  an  unknown  part  in  terms  of  a  given 
part,  and  a  part  already  found. 

Case  1.     By  §  18,  tan  <£  ==  ctn  (90' 


Also,  r  =  +Vx2-fy2. 

But  in  order  to  use  logarithms  we  take  r  =  x  sec  $  =  y  esc  <£. 

Case  2.     By  §  18,  cos  <f>  =  sin  (90°  —  <£)  =  -• 


Also,  y  =  +  Vr2  -  x2  =  +  V(r  +  x)  (r  —  x). 

Slog  y  =  *  [log  (r  +  x)  +  log  (r-x)];. 

If  y  is  given,  the  acute  angles  and  x  can  be  found  in  a 
similar  way. 

Case  3.     By  §  18,         x  =  r  cos  tj>. 
y  =  r  sin  <£. 

Solve  similarly  if  the  other  acute  angle  is  given. 

Case  4.     By  §  18,          r  —  x  sec  <£. 

y  =  x  tan  <£. 

Solve  similarly  if  y  is  given. 

B.  Examples.    VII. 

Find  the  unknown  parts  of  the  following  right  triangles,  having  given : 

1.  x  =  273,  y=  115. 

2.  x  =  23,  y  =  7. 

3.  r  =  97,  x  =  16. 

4.  r  =  42,  y  =  37. 

5.  r  =  67,  0  =  43°  25'. 

6.  r  =  19,  90°  -  0  =  27°  17X. 

7.  x  =  17,  0  =  47°  32'. 

8.  y  =  211,  90°  -  0  =  47°  16'. 


54 


PLANE   TRIGONOMETRY. 


C.  Practical  Applications  to  Problems   on  Heights 
and  Distances. 


§  63.  The  angle  of  elevation,  or  de- 
pression, of  a  point  P  is  the  angle  between 
a  horizontal  line  and  the  line  of  vision  to 
the  point  P,  both  lines  being  taken  in  the 
same  vertical  plane.  In  the  accompany- 
ing figure  the  eye  of  the  observer  is  sup- 
posed to  be  at  0.  In  the  case  of  a  heav- 
enly body  its  angle  of  elevation  is  called 
its  altitude.  In  other  cases  altitude  means 
the  height  of  a  point.  Many  problems  on 
heights  and  distances  can  be  made  to 
depend  upon  the  solution  of  right  tri- 
angles. 


FIG.  28. 


/>.   Examples.    VIII. 

1.  When  the  sun's  altitude  is  30°,  the  length  of  the  shadow  of  Bunker 
Hill  monument  on  the  horizontal  plane  is  381.1  ft.      How  high  is  the 
monument  ? 

2.  At  the  distance  of  75  ft.  from  its  base,  the  angle  of  elevation  of  the 
top  of  a  tree  standing  on  a  horizontal  plain  is  63°  26'.     How  high  is  the 
tree? 

3.  A  steeple  is  120  ft.  high.     What  is  the  angle  of  elevation  of  its  top 
at  a  distance  of  75  ft.  from  its  base  ? 

4.  A  man  on  the  top  of  a  tower  250  ft.  high  observes  an  object  whose 
angle  of  depression  is  53°  27'.     How  far  is  the  object  from  the  base  of  the 
tower  ? 

5.  At  two  points,  75  yards  from  each  other,  in  a  horizontal  line,  and 
in  the  same  vertical  plane  with  the  top  of  a  hill,  the  angles  of  elevation 
are  observed  to  be  45°  and  60°.     Find  the  height  of  the  hill  and  the 
distance  from  the  nearer  point  to  the  point  directly  under  the  top  of 
the  hill. 

6.  A  besieging  party  comes  to  a  channel  at  a  point  directly  opposite  a 
fortified  castle  of  the  enemy.     They  measure  away  from  the  shore,  and 
in  a  line  with  the  castle,  500  yards.     The  angles  of  elevation  of  the  top 
of  the  castle  are  observed  to  be  16'  15",  and  15'  50"  at  the  extremities  of 
the  base  line  respectively.     How  high  is  the  castle  ?     If  the  maximum 


EXAMPLES.     VIII.  55 

range  of  their  guns  is  ten  miles,  will  there  be  any  use  in  their  trying  to 
storm  the  castle  ?     (log  esc  25"  =  3.9105.) 

7.  From  the  top  of  a  tower  175  ft.  high  two  objects  on  the  horizontal 
plane,  in  a  line  with  the  base  of  the  tower,  are  observed  to  have  angles  of 
depression  of  60°  and  70°  respectively.     What  are  their  distances  from 
the  foot  of  the  tower,  and  from  each  other  ? 

8.  At  two  points,  a  feet  from  each  other,  in  a  horizontal  line,  and  in 
the  same  vertical  plane  with  the  top  of  a  tower,  the  angles  of  elevation 
are  observed  to  be  a  and  /3,  respectively,  where  a  >  /3.     Find  the  height, 
h,  of  the  tower,  and  the  distance,  b,  from  the  nearer  point  of  observation 
to  the  base  of  the  tower. 

Ans.    h  =  a  sin  a  sin  p  esc  (a  if  /3),  b  =  a  cos  a  sin  /3  esc  (a  q=  /9),  accord- 
ing as  the  points  are  on  the  same,  or  on  opposite  sides  of  the  tower. 

9.  Use  the  formulas  obtained  in  Problem  8  to  solve  Problems  5  and  6. 

10.  From  the  top  of  a  tower  h  ft.  high  two  objects  in  the  horizontal 
plane,  in  a  line  with  the  base  of  the  tower,  are  observed  to  have  angles  of 
depression  of  a  and  /3,  respectively,  where  a  >  /3.     What  are  their 
distances  from  the  foot  of  the  tower,  and  from  each  other  ? 

11.  A  farmer  wishes  to  make  a  ditch  100  rds.  long  on  level  land.     If 
the  inclination  of  the  bottom  of  the  ditch  to  the  horizon  is  12',  and  the 
depth  of  the  upper  end  of  the  ditch  is  12  in. ,  what  is  the  depth  of  the 
lower  end  ? 

12.  A  winding  railroad  is  built  around  a  cylindrical  tower  whose 
radius  is  100  ft.  and  height  500  ft.     If  the  maximum  inclination  to  the 
horizon  which  can  safely  be  used  is  10°,  how  many  times  will  the  road 
wind  around  the  tower  ?     If  the  road  is  3  ft.  wide,  what  is  the  length  of 
a  line  midway  between  the  rails  ? 

13.  The  length  of  a  road  (ascent  1  ft.  in  5),  from  the  foot  to  the  top 
of  a  hill,  is  If  mi.      What  will  be  the  length  of  a  zigzag  road  (ascent  1  ft. 
in  12)  ? 


CHAPTER   VII. 


A.    Some   General  Formulas  for  Oblique 
Triangles. 

§  64.  In  treating  the  oblique  triangle  we  limit  the  dis- 
cussion to  ^  geometrical  triangles.  We  shall  denote  the  sides 
of  a  triangle  by  a,  b,  c,  and  the  angles  opposite  these  sides  by 
A,  B,  C,  respectively. 

a.   Theorem  of  Sines. 

§  65.   In  any  plane  triangle  the  sides  are  proportional  to 

the  sines   of  the  opposite 
angles. 

Let  ABC  represent  any 
plane  triangle.  Let  p  rep- 
resent a  perpendicular  let 
fall  from  any  vertex  upon 
the  opposite  side. 

By  §  18,  p  =  b  sin  A, 

p  =  a  sin  B, 
.'.  b  sin  A  =  a  sin  B. 

Similarly,  c  sin  B  =  b  sin  C. 

a  sin  C  =  c  sin  A. 

These  equations  may  be  written : 

a  b  c 

sin  B  ~~ 


or 


sin  A      sin  B      sin  C 
a  :  b  :  c  I :  sin  A  :  sin  B  :  sin  C. 


GENERAL  FORMULAS  FOR   OBLIQUE   TRIANGLES.         57 


6.   Theorem  of  Tangents. 


§66.    From 


sin  A 
sin  B 


a+  b  _  sin  A  +  sin  B  _  tan  j-  (A  +  B) 
'a— b      sin  A  — sin  B       tan-j-(A  —  B) 


by  §  60. 


Similarly, 


b  — ctan£(B  — C) 
c  +  a  =  tan  j-  (C  +  A) 
c  —  a  tan  $  (C  —  A) 


c.   Theorem  of  Cosines. 

§  67.    Formulas  for  one  side  of  a  triangle,  in  terms  of  the 
other  two  sides,  and  their  included  angle. 

C 


FIG.  30. 
Whatever  allowable  value  of  A  is  used, 

v 

-  =  cos  A,  or  x  —  b  cos  A. 
b 


Similarly, 


=  y2  +  (c  -  x)2 

=  f  +  x2  +  c2  -  2  ex 

=  b2  +  c-  -  2  be  cos  A. 

b2  =  c2  -f-  a2  —  '2  ca  cos  B. 

C2  =  a2+  b2  —  2ab  cos  C 


58 


PLANE   TRIGONOMETRY. 


d.   Formulas  for  one  Side  of  a  Triangle,  in  Terms  of  the 
Adjacent  Angles,  and  the  other  two  Sides. 


FIG.  31. 


§68.   By  §53, 


or 


Similarly, 


a  =  c  cos  B  +  b  cos  (360°  —  C). 
a  =  b  cos  C  +  c  cos  B. 
b  =  c  cos  A  +  a  cos  C. 
c  —  a  cos  B  +  b  cos  A. 


ABC 
e.  Formulas  for  the  Functions  of  -,  — ,  —  • 


.   By  §  67,        cos  A  = 


b2  +  c2-; 
2  be 


sin     A  = 


1- 


b2+c2-a2 

2  be         by  §  58, 


2 


_Ja2-(b-e)2 
4  be 


_J(a-b  + 


b-c). 


4  be 


Put 
Then, 


s  = 


a+b 


-a+b+c 


a- b  +  c 


GENERAL  FORMULAS  FOR  OBLIQUE   TRIANGLES.        59 
a  +  b-c 


s  —  c 


sin 


_      2(s-b)2(s-c) 
y  4bc 

.  -  J(s-bxr^). 

*  be 


Similarly,  sin  £  B  =  + 


By  §58, 


2  be 


4  be 


=+v^- 

Similarly,  cos  £  B  =  +  A/S  (s  ~  b)  • 

ca 


cos  4-  C  = 


Since        tan     A  = 


cos     A 


Similarly,  tan  *  B  =  +      ($      c)  ($,^  a) 

s (s  —  b) 


60 


PLANE   TRIGONOMETRY. 


f.  Formulas  for  the  Area  of  a  Triangle,  and  the  Radii  of  the 
Inscribed  and  Circumscribed  Circles. 


§  7O.   By  geometry, 
area  of  the  triangle. 
But, 


By  §  65, 


represents  the 


y  =  b  sin  A. 
=  £  be  sin  A. 
a  sin  B 


b  = 


c  = 


sin  A 
a  sin  C 


sin  A 
a2  sin  B  sin  C       ,  a2  sin  B  sin  C 


GENERAL  FORMULAS  FOR   OBLIQUE   TRIANGLES. 


61 


Let  0  be  the  center  of  the  inscribed  circle  whose  radius  is  r. 
By  geometry,  AM  =  AP  [=  a']. 

BM  =  BN[=bf]. 


c'+a'  =  b. 
a'-fb'  =  c. 

Adding  and  dividing  by  2, 
a'+b'  +  c'  =  s. 

.-.  a'=  s  —  a. 
b'=s-b. 
c'  =  s-c. 

A          r 
.*.  tan  —  = 


2      s  —  a 

B  r 

tan—  = 


2       s  —  b 

C  r 

tan  -  = 


2       s  —  c 


.-.  4. 


=  rs  =  Vs  (s  -  a)  (s  -  b)  (s  -  c). 

A 

b  , 


Let  P  be  the  center  of  the  circumscribed  circle  whose 
radius  is  R.  By  geometry  we  have  the  parts  as  designated 
in  the  figure. 


62  PLANE  TRIGONOMETRY. 


a=B. 


.-.a  =  90°  —  A. 

a. 

2 

cos  a  =  sin  A  =  -ir 
R 

.-.  R=|cscA. 


be 
By  1,  esc  A  = 


5.   R=  abc 


This  demonstration,  in  connection  with   §  65,  gives  inci- 
dentally the  interesting  relation 


sin  A      sin  B      sin  C 

The  formulas  of  this  chapter  enable  us  to  solve  any  plane 
triangle  when  the  necessary  parts  are  given,  and  also  to  find 
its  area. 


B.  Oblique  Triangles. 
a.   Cases. 

§71.    There  are  four  cases. 

1.  Given  two  angles  and  a  side. 

2.  Given  two  sides  and  an  angle  opposite  one  of  them. 

3.  Given  two  sides  and  the  included  angle. 

4.  Given  three  sides. 


OBLIQUE   TRIANGLES. 


63 


b.   Solutions. 

Case  1.     The  third  angle  is  the  supplement  of  the  sum  of 
the  two  given  angles.*    The  required  sides  are  given  by  the 
formulas  of  §  65.     For  example,  if  a  is  the  given  side, 
b  =  a  sin  B  esc  A.  c  =  a  sin  C  esc  A. 

Case  2.     If  a,  b,  and  A  are  given, 

,m  B  =  ^^  by  §66. 


Since  B  is  found  from  its  sine,  and  since  there  are  always 
two  angles,  each  less  than  180°,  which  have  the  same  sine, 
B  may  have  two  values,  and  the  triangle  may  have  two 
solutions.  But  if  A  is  obtuse,  there  cannot  be  more  than  one 
solution.  The  cases  may  be  discriminated  as  follows  : 

(1)  A  acute. 

(1)  Two  solutions  when 
b  >  a  >  b  sin  A. 

(2)  One   solution    when 
a  >  b,  a  =  b  (isosceles), 

a  =  b  sin  A  (right  angled). 

(3)  No  solution  when 
a  <  b  sin  A. 

(#)   A  obtuse. 

(1)  One  solution  when  a  >  b. 

(2)  No  solution  when     b  ^>  a. 
Case  8.     Given  a,  b,  and  C. 

(A-B)  =  a~b 


FIG.  35. 


By  §66,          tan 


. 
a  -)-  b 

Since,  tan  \  (A  +  B)  =  ctn  %  C. 

This  gives      tan  J  (A  —  B),  and  therefore  |  (A  —  B). 
Then,  i(A+B).+  i(A-  B)  =  A. 

i(A+B)-f(A-B)=B. 


64 


PLANE   TRIG  GNOME  TR  Y. 


The  required  side  can  now  be  found  by  §  65. 

Case  4'     Given  a,  b,  and  c. 

By  §  69  or  §  70  the  angles  can  be  found. 

The  tangent  formulas  should  be  used  in  practice. 


FIG.  36. 


C.  Examples.    IX. 

1.  A  =  67°  24',   B  =  39°  18',  a  =  15. 

2.  C  =  25°  19',  A  =  74°  30',  c  =  23. 

3.  a  =  90,  b  =  100,  A  =  CO0. 

4.  b  =  65,  a  =  124,  A=  117°. 

5.  c  =  96,  b  =  45,     C  =  30°. 

6.  a  =  25,  b  =  19,     C  -  59°. 

7.  c  =  43,  b  =  29,     A  =  40°. 

8.  a  =  25,  b  =  33,     c  =  44. 

9.  a  =  37,  b  =  14,     c  =  29. 

10.  a  =  29,  b  =  57,     c  =  27. 

11.  Are  the  following  triangles  possible?     If  impossible,  what  con- 
ditions are  violated  ?  "  If  possible,  have  they  one  or  two  solutions  ? 


EXAMPLES.     IX.  65 

(1)  A  =  95°  27',   B  =  84°,  35',  a  =  73. 

(2)  a  =  47,   b  =  53,  A  =  91°  16'. 

(3)  b  =  27,  c  =  54,   B  =  30°. 

(4)  a  =  65,  b  =  70,  A  =  58°. 

(5)  a  =  34,  b  =  79,  c  =  44.5. 

12.  The  angle  of  elevation  of  a  balloon  which  is  ascending  uniformly 
and   vertically,   when   it  is  one   mile  high  is  observed  to  be  35°  20'; 
20  minutes  later  the  elevation  is  observed  to  be  55°  40'.     How  fast  is  the 
balloon  moving  ? 

13.  A  base  line,  AB,  1000  ft.  long  is  measured  along  the  straight  bank 
of  a  river ;  C  is  an  object  on  the  opposite  bank  ;  the  angles  BAG  and  CBA 
are  observed  to  be  65°  37'  and  53°  4'  respectively.     Find  the  breadth  of 
the  river  at  C. 

14.  A  column  on  a  pedestal  20  ft.  high  subtends  an  angle  45°  to  a 
person  on  the  ground ;  on  approaching  20  ft.  it  again  subtends  an  angle 
45°.     Find  the  height  of  the  column. 

15.  A  and  B  are  two  points  on  opposite  sides  of  a  swamp,  and  C  is  a 
point  visible  from  both  A  and  B.     AC  is  15  rods,  and  BC  12  rods;  the 
angle  ACB  is  60°.     Find  the  distance  AB. 

16.  Two  mountains,  each  one  mile  high,  are  just  visible  from  each 
other  over  the  sea.     If  the  radius  of  the  earth  is  taken  equal  to  4000  miles, 
how  far  are  the  mountains  apart  ? 

17.  Two  men  standing  at  the  same  point,  C,  observe  the  horizontal 
angle  subtended  by  the  line  joining  two  inaccessible  objects,  A  and  B ; 
they  then  move  away,  one  in  the  direction  AC  to  D,  the  other  in  the 
direction  BC  to  E,  until  each  observes  the  horizontal  angle  to  be  half 
what  it  was  before.     ACB  =  30°,  CD  =  100,  CE  =  200.     Determine  AB. 

18.  A  base  line  400  yards  long,  whose  upper  end  is  8  yards  higher  than 
the  lower,  is  measured  in  the  same  vertical  plane  with  the  top  of  a  hill. 
The  angles  of  elevation  of  the  top  of  the  hill,  measured  at  the  lower  and 
upper  ends  of  the  base  line,  are  found  to  be  5°  17'  and  3°  17'  respectively. 
Find  the  height  of  the  hill. 

19.  A  pole  100  ft.  high  stands  in  the  center  of  a  horizontal  equilateral 
triangle.     At  the  top  of  the  pole  each  side  subtends  an  angle  60°.     Find 
the  length  of  a  side  of  the  triangle. 

20.  A  regular  pyramid  on  a  square  base  has  an  edge  of  150  ft.  in 
length.     The  side  of  the  base  is  200  ft.  long.     Find  the  inclination  of  the 
face  to  the  base. 

21.  The  distance  between  two  lighthouses,  one  of  which  bears  E  from 
the  other,  is  4  miles.     The  bearings  of  the  lighthouses  from  a  ship  are 


66  PLANE  TRIGONOMETRY. 

E.b.N.  and  N.W.b.W.  respectively.     What  is  the  distance  of  the  ship 
from  each  lighthouse  ? 

22.  Two  cruisers  start  from  the  same  point,  and  steam  at  the  rate  of 
19  and  23  knots  respectively,  one  S. E.,  and  the  other  S.|W.     How  far 
apart  are  they  in  five  hours  ? 

23.  A  man  standing  on  a  horizontal  plain  at  a  distance  a  feet  from  a 
tower  observed  that  a  flagstaff  on  the  tower  subtends  an  angle  a,  and  that 
on  walking  2  b  feet  towards  the  tower  the  flagstaff  again  subtends  the 
same  angle.     Prove  that  the  height  of  the  flagstaff  is  2  (a  —  b)  tan  a  feet. 

24.  A  tower  and  a  spire  on  its  top  subtend  equal  angles  at  a  point 
whose  distance  from  the  foot  of  the  tower  is  a.     If  h  is  the  height  of  the 

tower,  prove  that  the  height  of  the  spire  is  h    2  _      • 

25.  A  lighthouse  is  N.b.W.  of  a  ship.     After  the  ship  sails  21  miles 
E.S. E.,  the  bearing  of  the  lighthouse  is  N.W.     What  is  the  distance  of 
the  lighthouse  from  each  position  of  the  ship  ? 

26.  Find  the  area  of  a  quadilateral  field  whose  sides,  AB,  BC,  CD,  are 
20,  15,  and  18  respectively,  and  whose  angles,  B  and  C,  are  85°  and  116° 
respectively. 


SPHERICAL    TRIGONOMETRY. 

CHAPTER   I. 
A.   Introduction. 

a.   Geometrical  Principles, 

§  72.  1.  Every  tangent  to  a  sphere  is  perpendicular  to  the 
radius  drawn  to  the  point  of  contact,  and  is  also  tangent  to 
that  great  circle  arc  in  whose  plane  it  lies. 

2.  Two  angles  in  space  which  have  their  sides  parallel 
respectively,  and  lying  in  the  same  or  in  opposite  directions 
from  their  vertices,  are  equal. 

ABC 
In  any  spherical  triangle  : 

clDC 

3.  A  =  7T-a',         B  =  9T-bf,        C  =  ir  —  C', 

a  =  Tr—  A',     b=?r—  B',     C  =  TT  —  C', 

A'B'C'  .  f  ABC 

where  ,  is  the  polar  triangle  01 

a'b'c'  abc 

4.  The  greater  side  is  opposite  the  greater  angle,  and  con- 
versely. 

5.  The  sum  of  two  sides  is  greater  than  the  third. 
That  is,  a  +  b  >  c,  b  +  c  >  a,  c  +  a  >  b. 

6.  27T>7T+A>B  +  C, 

27r>7r+B>C+A, 


68 


SPHERICAL   TRIGONOMETRY. 


&.   Sufficiency  of  the  Convex  Spherical  Triangle. 

The  solution  of  every  spherical  triangle  may  be  made  to 
depend  upon  the  solution  of  a  triangle  in  which  the  parts  are 
each  separately  less  than  TT.  This  is  made  evident  by  the 
following  summary  : 


denote  that  k 


f  sides 
\  angles 


of  a  spherical  triangle  are  >  TT,  the  possible  combinations  are 
16,  of  which  8  are  excluded,  and  8  give  spherical  triangles  : 


siao 


5.   s2a0 


7.  S3a0 


2.  s0a3 


6.   S2a3 


.   S3a3 


These  eight  cases  are  represented  graphically  by  the  follow- 
ing three  figures.  The  numbers  accompanying  the  figures 
correspond  respectively  to  the  cases. 


1  &2 


FIG.  37. 


FIG.  38. 


SPHERICAL   TRIGONOMETRICAL  RELATIONS. 

5  &  6,  and  7  &.  8 


\ 


69 


FIG.  39. 


B.   General  Theory  of  Spherical  Trigonometrical 
Relations. 

§  73.  It  is  shown  in  geometry  that  a  spherical  triangle  is 
determined  if  any  three  parts,  not  excluding  the  three  angles, 
are  properly  given.  In  the  general  case,  the  three  parts  are 
independent  of  each  other,  and  relations  between  the  parts  of 
the  general  triangle  cannot,  therefore,  exist  unless  the  parts 
enter  into  such  relations  at  least  four  at  a  time.  For  example, 
if  c  — f(b),  or  if  c  =  f(a,  b),  c  is  determined,  and,  therefore, 
three  parts  are  determined  as  soon  as  a  and  b  are  given  in 
either  case ;  that  is,  the  triangle  is  determined  by  two  parts. 
But  if  A  — f(a,  b,  c),  A  is  not  determined  until  three  inde- 
pendent parts,  a,  b,  and  c,  have  been  given. 

§  74.  Let  us  inquire  how  many  species  of  relations  we 
may  expect  to  meet.  In  the  relations  of  parts  taken  four  at 
a  time  we  shall  pick  out  four  and  always  exclude  two  parts. 
The  number  of  species  of  parts  retained  will  be  the  same  as 
the  number  of  species  of  parts  excluded.  For  convenience,  it 
will  be  easier  to  consider  the  latter.  The  number  of  combina- 


6X5 

tions  of  six  things  taken  two  at  a  time  is =  15. 

1  X  ^ 


The 


70  SPHERICAL   TRIGONOMETRY. 

six  parts  of  the  triangle  are  a,  b.  c,  A,  B,  C.  These  fifteen 
combinations  are  ab,  ac,  aA,  aB,  aC,  be,  bA,  bB,  bC,  cA,  cB, 
cC,  AB,  AC,  BC.  Excluding  those  which  are  repetitions,  we 
find  only  four  distinct  species  represented  by  1.  BC,  2.  cC, 
3.  cB,  4.  be,  and  these  give  us,  therefore,  to  enter  into  rela- 
tions four  at  a  time  the  four  distinct  species  where  the  parts 
retained  are  the  complementary  combinations  of  the  parts 
excluded,  viz.: 

Ketained   1.  abcA,       2.  abAB,       3.  abAC,       4.  ABCa. 
Excluded  1.  BC,          2.  cC,  3.  cB,  4.  be. 

§  75.  To  find  the  number  of  species  of  parts  taken  five  at 
a  time,  we  may  proceed  as  above. 

The  number  of  combinations  of  six  things  taken  one  at  a 
time  is  six.  The  six  combinations  are  a,  b,  c,  A,  B,  C.  Here 
are  only  two  distinct  cases,  1.  C,  2.  c,  and  we  have : 

Eetained   1.  abcAB,         2.  abABC. 
Excluded  1.  C,  2.  c. 

When  the  parts  are  taken  six  at  a  time  there  can  be  only 
one  species  of  relations. 

C.   Systems  of  Equations  demanded  by  the 
Preceding  Theory. 

a.   Equations  in  which  the  Parts  enter  Four  at  a  Time. 

These  are  by  §  74  of  four  species. 

§  76.  1.  Species  one.  In  which  we  have  three  sides  and  an 
angle. 

Let  ABC  be  any  spherical  triangle.  Let  0  be  the  center  of 
the  sphere.  OA,  OB,  OC,  are  radii  of  the  sphere.  In  the 
plane  OAB  draw  DB  and  OM  perpendicular  to  AO.  AT,  the 
tangent  to  the  arc,  AB,  at  A  is  also  perpendicular  to  OA. 
Hence  AT,  DB,  and  OM  are  parallel  lines.  In  the  plane 


SYSTEMS  OF  EQUATIONS. 


71 


FIG.  40. 

OAC  draw  PC  and  ON  perpendicular  to  OA  ;  then  AT,  the 
tangent  to  the  arc  AC  at  A,  PC,  and  ON  are  parallel.  And 
the  angle  NOM  =  the  angle  T'AT  =  the  angle  A  of  the 
spherical  triangle  ABC. 

§77.   By  §53,          OBOC  =  ODOC  +  DBOC, 

.'.  OB  cos  a  =  OD  cos  b  -f  DB  cos  COM. 


And 

But 

hence 
or 

Also 


OC  cos  MOC  =  OC  cos  COM  =  0  +  PC  cos  A. 

PC  =  OC  sin  b, 

OC  cos  COM  —  OC  sin  b  cos  A, 
cos  COM  =  sin  b  cos  A. 

OD^OBcosc, 
DB  =  OB  sine. 


Substituting  these  values  in  the  equation, 

OB  cos  a  =  OD  cos  b  +  DB  cos  COM, 


72  SPHERICAL   TRIGONOMETRY. 

we  get  OB  cos  a  =  OB  cos  b  cos  c -f  OB  sin  b  sin  c  cos  A, 

or  cos  a  =  cos  b  cos  c  +  sin  b  sin  c  cos  A. 

Advancing  the  letters,  cos  b  =  cos  c  cos  a  +  sin  c  sin  a  cos  B, 
cos  c  =  cos  a  cos  b  +  sin  a  sin  b  cos  C. 

§  78.  Prom  these  three  equations  all  other  equations  what- 
ever, which  are  true  of  any  spherical  triangle,  may  be  obtained 
by  algebraic  transformations,  and  are  therefore  dependent  or 
derived  equations.  For  as  three  parts  of  a  spherical  triangle 
may  be  taken  arbitrarily,  but  when  they  are  taken  the  tri- 
angle is  fixed,  and  three  parts  must  be  taken  to  fix  the  triangle, 
there  remain  three  parts  unknown.  Now  the  number  of 
simultaneous  and  independent  equations  required  to  deter- 
mine three  unknown  quantities  is  three.  As  many  as  three 
must  be  given,  but  not  more  than  three  equations  can  exist 
independently.  If,  then,  we  have  a  system  or  systems  of 
simultaneous  equations  involving  three  unknown  quantities, 
not  more  than  three  of  the  equations  can  be  taken  to  be  inde- 
pendent, and  as  many  as  three  of  them  must  be  taken  to  be 
independent,  if  the  system  is  determinately  solvable.  From 
this  it  is  evident  that  we  might  take  any  one  of  the  systems 
of  three  equations,  where  the  three  equations  are  independent 
of  each  other,  as  the  three  fundamental  equations  of  spherical 
trigonometry.  But  we  will  choose  the  preceding  three  equa- 
tions of  §  77,  as  the  fundamental  equations.  We  shall  now 
show  how  the  other  systems  of  equations  are  derived  from 
them. 

2.  Species  two.  Equations  involving  two  sides  and  the  angles 
opposite  to  them. 

§  79.  Here,  e.g.,  we  must  retain  abAB,  and  exclude  cC. 
By  adding  and  subtracting  the  first  two  of  the  equations  of 
§  77,  we  have  : 

(cos  a  +  cos  b)  (1  —  cos  c)  =  sin  c  (sin  b  cos  A  +  sin  a  cos  B). 
(cos  a  —  cos  b)  (1  +  cos  c)  =  sin  c  (sin  b  cos  A  —  sin  a  cos  B). 


SYSTEMS   OF  EQUATIONS.  73 

Multiplying  these  two  equations  together,  and  dividing  by 
sin2c  =  1  —  cos2c,  which  we  have  a  right  to  do,  since  c  =%=  0, 
sin  c  =)=  0,  and  therefore  sin2c  =(=  0?  an(i  we  g6^  : 

cos2a  —  cos2b  —  sin2b  cos2 A  —  sin2a  cos2B, 
or  expressing  in  terms  of  sines, 

sin2b  sin2 A  —  sin2a  sin2B,     .'.  sin  b  sin  A  —  ±  sin  a  sin  B, 

but  as  all  the  factors  involved  in  the  equation  are  positive, 
we  must  exclude  the  minus  sign,  and  write : 

sin  b  sin  A  =  sin  a  sin  B. 
Advancing  the  letters, 

sin  c  sin  B  =  sin  b  sin  C, 
sin  a  sin  C  =  sin  c  sin  A. 

These  equations  may  also  be  put  in  the  forms : 

sin  a sin  b sin  c  _ 

sin  A  ~~  sin  B  ~~~  sin  C  ~~ 
sin  a  =  r  sin  A,         sin  b  =  r  sin  B,         sin  c  =  r  sin  C. 

3.  Species  three.  Equations  involving  two  sides  and  an  angle 
opposite  to  one  of  them. 

§  8O.    We  must,  e.g.,  retain  abAC,  and  exclude  cB. 

In  the  first  of  the  equations  of  §  77,  substitute  for  cos  c  its 
value  given  by  the  third  equation,  and  for  sin  c,  its  value, 

— : — -r — >  given  by  the  third  one  of  the  equations  of  §  79. 

We  get : 

cos  a  =  cos  b  (cos  a  cos  b  +  sin  a  sin  b  cos  C) 
sin  b  sin  a  sin  C  cos  A 

sin  A 

cos  a  (1  —  cos2b)  =  cos  a  sin2b  =  cos  b  sin  a  sin  b  cos  C 
+  sin  b  sin  a  sin  C  ctn  A. 


74  SPHERICAL   TRIGONOMETRY. 

Dividing  both  sides  of  the  equation  by  sin  a  sin  b,  since 
sin  b  sin  c  •=$=•  0, 

ctn  a  sin  b  =  cos  b  cos  C  +  sin  C  ctn  A. 
Advancing  the  letters, 

ctn  b  sin  c  ^  cos  c  cos  A  +  sin  A  ctn  B, 
ctn  c  sin  a  =  cos  a  cos  B  +  sin  B  ctn  C. 

If  instead  of  retaining  abAC,  and  excluding  cB,  we  had 
retained  acAB,  and  excluded  bC,  we  should  simply  have  put 
c  in  the  place  of  b,  and  B  in  the  place  of  C,  and  instead  of 
the  first  of  the  preceding  equations,  we  should  have  had : 

ctn  a  sin  c  =  cos  c  cos  B  +  sin  B  ctn  A. 

Advancing,     ctn  b  sin  a  =  cos  a  cos  C  '-{-  sin  C  ctn  B, 

ctn  c  sin  b  =  cos  b  cos  A  +  sin  A  ctn  C. 

4.   Species  four.    Equations  involving  three  angles  and  a  side. 

§  81.   Here  we  must,  e.g.,  retain  ABCa,  and  exclude  be. 
If  we  divide  the  equation  at  the  beginning  of  the  preceding 
section  by  sin  b  instead  of  sin  a  sin  b,  there  results  : 

cos  a  sin  b  =  cos  b  sin  a  cos  C  +  sin  a  sin  C  ctn  A.          « 

-01  r>  i,  .  sin  a  sin  C 

Keplacmg  sin  b  by  r  sin  B,  sm  a  by  r  sin  A,  sm  c  = : — r — 

sin  A 

by  r  sin  C,  and  dividing  by  r,  we  have  : 

cos  a  sin  B  =  cos  b  sin  A  cos  C  +  sin  C  cos  A. 

If  cos  c  is  replaced  in  the  second  of  the  equations  of  §  77 
by  its  value  taken  from  the  third,  the  result  will  differ  from 
the  preceding  in  having  a  changed  into  b,  and  A  into  B,  and 
it  will  be : 

cos  b  sin  A  =  cos  a  sin  B  cos  C  +  sin  C  cos  B. 

Multiplying  the  latter  equation  by  cos  C,  and  equating  the 
terms  cos  b  sin  A  cos  C,  there  results  : 


SYSTEMS  OF  EQUATIONS.  75 

cos  a  siri  B  —  sin  C  cos  A  =  cos  a  sin  B  cos2C  +  sin  C  cos  C  cos  B, 
or         cos  a  sin  B  (1 ' —  cos2C)  =  cos  a  sin  B  sin2C 

=  sin  C  cos  A  +  sin  C  cos  C  cos  B. 

Dividing  by  sin  C,     since  sin  C  =(=  0, 

cos  a  sin  B  sin  C  =  cos  A  +  cos  B  cos  C, 
or  cos  A  =  —  cos  B  cos  C  +  sin  B  sin  C  cos  a. 

Advancing,     cos  B  =  —  cos  C  cos  A  +  sin  C  sin  A  cos  b, 
cos  C  —  —  cos  A  cos  B  +  sin  A  sin  B  cos  c. 

b.   Equations  in  which  the  Parts  enter  Five  at  a  Time. 

1.  Species  one.  Equations  involving  three  sides  and  two 
angles. 

§  82.   We  retain  here,  e.g.,  abcAB,  and  exclude  C. 

If  in  the  first  of  the  equations  of  §  77,  we  substitute  for 
cos  b,  its  value  as  given  by  the  second  of  those  equations, 
cos  a  =  (cos  c  cos  a  +  sin  c  sin  a  cos  B)  cos  c  +  sin  b  sin  c  cos  A, 
or  cos  a  (1  —  cos2c)  =  cos  a  sin2C  =  sin  c  sin  a  cos  c  cos  B 

+  sin  b  sin  c  cos  A. 

Dividing  by  sin  c,     since  sin  c  =|=  0, 

cos  a  sin  c  =  cos  c  sin  a  cos  B  +  sin  b  cos  A, 
or  as  usually  written, 

sin  b  cos  A  =  cos  a  sin  c  —  cos  c  sin  a  cos  B. 

Advancing,      sin  c  cos  B  =  cos  b  sin  a  —  cos  a  sin  b  cos  C, 
sin  a  cos  C  =  cos  c  sin  b  —  cos  b  sin  c  cos  A. 

If  instead  of  putting  cos  b  in  the  first  of  the  equations  of 
§  77,  we  had  put  it  in  the  third,  the  results  would  have 
appeared  with  a  and  c,  and  A  and  C,  interchanged,  or 

sin  b  cos  C  =  cos  c  sin  a  —  cos  a  sin  c  cos  B, 

and  also,  sin  c  cos  A  =  cos  a  sin  b  —  cos  b  sin  a  cos  C, 

sin  a  cos  B  =  cos  b  sin  c  —  cos  c  sin  b  cos  A. 


76  SPHERICAL   TRIGONOMETRY. 

2.   Species  tivo.  Equations  involving  three  angles  and  two  sides. 
§  83.    Here  we  retain,  e.g.,  ABCab,  arid  exclude  c. 
The  first  of  these  equations  has  already  been  obtained  in 
§  81.     It  is  : 

cos  a  sin  B  =  cos  b  sin  A  cos  C  +  sin  C  cos  A. 

Advancing,      cos  b  sin  C  =  cos  c  sin  B  cos  A  +  sin  A  cos  B, 
cos  c  sin  A  =  cos  a  sin  C  cos  B  +  sin  B  cos  C. 

The  leading  equation  of  the  second  division  of  this  set  has 
also  been  obtained  in  §  81.     It  is : 

cos  b  sin  A  =  cos  a  sin  B  cos  C  +  sin  C  cos  B. 

Advancing,      cos  c  sin  B  =  cos  b  sin  C  cos  A  -+-  sin  A  cos  C, 
cos  a  sin  C  =  cos  c  sin  A  cos  B  +  sin  B  cos  A. 

c.   Equations  in  which  the  Parts  enter  Six  at  a  Time. 

§  84.   Here  there  is  but  one  species.      An  example  may 
suffice.     From  §  77  we  can  obtain, 

cos  a  cos  b  cos  c  —  (cos  b  cos  c  +  sin  b  sin  c  cos  A) 
(cos  c  cos  a  +  sine  sin  a  cos  B) (cos  a  cos  b-hsinasin  bcosC)=0, 
cos  A  cos  B  cos  C  sin2a  sin2b  sin2c  =  (cos  a  —  cos  b  cos  c) 
(cos  b  —  cos  c  cos  a)  (cos  c  —  cos  a  cos  b). 

D.  On  the  Use  of  the  Polar  Triangle  in  Establishing 

Relations. 

§85.    The  relations  A  =  TT  — a',    a  =  TT  —  A', 

B  =  7r-b',    b  =  7r-B',  where    ,^,, 

C  =  7T-C',       C-7T-C', 

is  the  polar  triangle  of      ,     ,  often  afford  an  easy  means  of 

a  b  c 

establishing  a  relation.     For  example,  we  may  say  that  the 
formulas  of  §  77  are  transformed  into  those  of  §  81.     This  is 
shown  as  follows  :    Express  that  the  formulas  of  §  77  hold 
for  the  polar  triangle.     Then  the  first  one  gives : 
cos  a'  =  cos  b'  cos  c'  +  sin  b'  sin  c'  cos  A'. 


USE   OF   THE  POLAR   TRIANGLE.  77 

Substituting  for  a',  b',  c',  A',  the  values  given  above,  and  we 
have  : 

cos  (TT  —  A)  =  cos  (TT  —  B)  cos  (TT  —  C)  +  sin  (TT  —  B) 

sin  (TT  —  C)  cos  (TT  —  a), 

or  —  cos  A  =  ( —  cos  B)  ( —  cos  C)  +  sin  B  sin  C  ( —  cos  a), 

or  cos  A/=  —  cos  B  cos  C  +  sin  B  sin  C  cos  a. 

The  equations  of  §  79  are  transformed  into  themselves ; 
that  is,  we  get  nothing  different  by  using  the  polar  triangle 
with  them.  The  first  half  of  the  formulas  of  §  80  are  trans- 
formed into  the  second  half,  and  the  second  half  into  the  first 
half.  Thus  the  first  one  of  the  first  half, 

ctn  a  sin  b  =  cos  b  cos  C  +  sin  C  ctn  A, 
is  transformed  into 

—  ctn  A  sin  B  =  cos  B  cos  c  —  sin  c  ctn  a, 
or  ctn  a  sin  c  =  cos  c  cos  B  +  sin  B  ctn  A, 

which  is  the  first  one  of  the  second  half.  The  formulas  of 
§  82  are  transformed  into  those  of  §  83,  and  vice  versa,  as  the 
student  may  satisfy  himself,  by  writing  out  the  leading  equa- 
tions and  making  the  substitutions.  As  this  work  can  be 
done  at  a  glance  when  once  understood,  it  is  evident  that  the 
method  affords  an  easy  and  powerful  means  of  writing  down 
new  formulas  without  any  work  at  all.  It  is  a  method  that 
ought,  at  least,  to  be  applied  to  every  new  formula  obtained, 
for  it  may  enable  one  to  double  the  number  of  his  formulas 
without  great  labor. 

GENERAL  NOTE  ON  SYSTEMS  or  TRIGONOMETRIC  EQUATIONS.  —  The 
preceding  investigations  have  shown  that  the  number  of  species  of  trigo- 
nometric equations  or  formulas  is  exactly  determinable.  And  while 
complete  sets  of  the  various  species  have  been  given,  it  is  to  be  observed 
that  other  sets  might  be  derived,  and  that  the  number  of  derived  sets  of 
the  various  species  is  unlimited. 


CHAPTEK   II. 

/.    Formulas   for   Spherical   Right,   and    Quadrantal 
Triangles. 

§  86.  In  the  formulas  of  §§  77-81,  making  an  angle,  and  a 
side,  as  C,  and  c,  successively  equal  to  90°,  we  have  the 
following  formulas  for  spherical  right,  and  quadrantal 
triangles  : 


a.   Spherical  Right  Trian 

1.  cos  c  =  oos  a  cos  b. 

0          •         A            Silia 

gles. 

§  77 

§  79 
§  79 
§80 
§  80 
§80 
§  80 
§81 

§81 
§  81 

b.   Quadrantal  Triangle 

1'.  cos  C  =  —  cos  A  cos  B. 
0.     .            sin  A 

JS. 

§81 
§  79 

§  79 
§  80 
§  80 
§80 
§80 
§77 

§  77 
§77 

25,   Sill  A  —    . 
sin  c 

»     -in  P       Slnb- 

A  .  bin  a  —    .     _ 
sin  C 

0.     .            sin  B 

Qf    Qin  h  —  -•             »               « 

O.     &111    D  . 

sin  c 
tan  b 

sin  C 
tanB 

4.  cos  A  —  . 
tan  c 

„           D      tan  a 

*±  .    CUO   d  ^ 

tanC 
tan  A 

o.  cos  b  —  . 
tan  c 

.       tan  a 

tanC 
tan  A 

o.  tan  A  —    .     ,  • 
sin  b 

7    1    nP      tanb- 

u  .  tan  a  —   .    -- 
sin  B 

tanB 

4  .     Lclll  L>  . 

sin  a 
cos  A 

sin  A 

.       cos  a 

o.  cos  a  —    .    _.  * 
sin  B 

cos  B 

o  .  Cub  A  —     .       . 

sm  b 
cos  b 

y.  cos  D  —    .     . 
sin  A 

10.  cos  c  =  ctn  A  ctn  B. 

y  .  cob  D  —    . 
sin  a 

10'.  cos  C  =  —  ctn  a  ctn  b. 

SPHERICAL  RIGHT  AND   QUADRANTAL   TRIANGLES.       79 


c.   Attention  is  called  to  the  Following  Deductions 
from  the  Formulas  of  §  86. 

§  87.  1.  From  1  and  1'  it  appears  that  c  will  be  of  the  first 
or  second  quadrant,  and  C  of  the  second  or  first  quadrant, 
according  as  a  and  b,  or  A  and  B,  respectively,  are  of  the 
same  or  of  different  quadrants. 

2.  From  6  and  7  it  appears  that  an  oblique  angle  and  its 
opposite  side,  and  from  6'  and  7'  that  a  non-quadrantal  side, 
and  its  opposite  angle,  are  of  the  same  quadrant. 


B.   Solution  of  Spherical  Right  and  Quadrantal 
Triangles. 

§  88.  In  the  right  spherical  triangle,  the  right  angle,  and 
in  the  quadrantal  triangle,  the  quadrant,  counts  as  one  of  the 
three  given  parts.  There  are  six  cases  to  be  treated. 


a.   Right  Triangle. 

b.   Quadrantal  Triangle. 

Case. 

Given. 

Re- 
quired. 

Formulas 
§  86.  a. 

Check. 

Case. 

Given. 

Re- 
quired. 

Formulas 
§86.  b. 

Check. 

1 

ab 

CAB 

1,6,7 

10 

1 

AB 

Cab 

l',6',7' 

10' 

2 

ac 

bAB 

1,2,5 

9 

2 

AC 

Bab 

l',2',5' 

9' 

3 

aA 

bcB 

6,2,8 

3 

3 

Aa 

BCb 

0',  2',  8' 

3' 

4 

aB 

bcA 

7,5,8 

4 

4 

Ab 

BCa 

7',  6',  8' 

4' 

5 

cA 

abB 

2,  4,  10 

7 

5 

Ca 

ABb 

2',  4',  10' 

r 

6 

AB 

abc 

8,  9,  10 

1 

6 

ab 

ABC 

8',  9',  10' 

V 

§  89.  The  remarks  of  §  87,  taken  in  connection  with  the 
formulas  given  in  §  88  for  the  solution  of  the  various  cases, 
show  that  of  the  six  cases,  only  the  third  is  ambiguous.  In 
this  case  the  required  parts  must  be  obtained  from  their  sines. 
Since  eacli  required  part  has  two  values,  supplements  of  each 


80  SPHERICAL   TRIGONOMETRY. 

other,  there  might  apparently  be  eight  triangles  ;  but  as  b 
and  B  (for  the  right  spherical  triangle)  must  be  of  the  same 
quadrant,  only  two  combinations  of  a,  b,  and  B  are  allowable, 
and  when  a  and  b  are  determined,  c  is  determined,  and,  there- 
fore there  cannot  be  more  than  two  solutions.  If  a  differs 
less  than  A  from  90°,  the  triangle  is  impossible,  since  the 
sine  of  each  required  part  would  be  greater  than  unity.  An 
analogous  consideration  holds  for  the  corresponding  case  of 
the  quadrantal  triangle. 

§  9O.  A  complete  set  of  formulas  could  be  deduced  for 
spherical  isosceles  triangles  from  the  general  formulas  of 
§§  77-81,  by  assuming  that  two  sides  are  equal,  but  as  they 
can  be  easily  solved  by  the  formulas  for  right  triangles,  we 
will  not  give  a  special  set  of  formulas. 


C.   Some  Astronomical  Conceptions. 

a.    Celestial  Sphere. 

1.  Definition.     The  Celestial  Sphere  is  a  sphere  of  indefi- 
nitely great  radius,  concentric  with  the  earth,  and  in  whose 
surface,    the   heavenly   bodies    are,    for   astronomical    uses, 
supposed  to  be. 

Its  axis,  called  the  Celestial  Axis,  coincides  with  the  axis 
of  the  earth,  and  its  ends  are  the  North  and  South  Poles  of 
the  celestial  sphere. 

2.  Great  Circles. 

(1)  The  Sensible  Horizon  is  that  small  circle  of  the  celes- 
tial sphere,  which  is  tangent  to  the  earth  at  the  observer's 
standpoint. 

The  Rational  Horizon  is  that  great  circle  of  the  celestial 
sphere,  which  is  parallel  to  the  sensible  horizon. 

Because  the  radius  of  the  celestial  sphere  is  indefinitely 
great,  we  are  justified,  in  practice,  in  identifying  the  sensible 


CELESTIAL  SPHERE.  81 

and  the  rational  horizon,  and  the  observer's  standpoint  and 
the  center  of  the  celestial  sphere.  Accordingly  we  shall, 
without  distinction,  speak  only  of  the  horizon. 

The  upper  and  lower  poles  of  the  horizon  are  called  the 
Zenith  and  Nadir,  respectively. 

(2)  Vertical  Circles  are  the  great  circles  which  pass  through 
the  zenith  and  nadir. 

(3)  The  Equinoctial,  or  Celestial   Equator,   is  that  great 
circle  of  the  celestial  sphere  which  is  perpendicular  to  the 
celestial  axis.     Obviously  its  poles  are  the  celestial  poles. 

(4)  Hour  Circles,  or  Declination  Circles,  are  great  circles 
passing  through  the  celestial  poles. 

(5)  The  Celestial  Meridian  is  that  hour  circle  which  passes 
through  the  zenith  and  nadir.     It  coincides  with  the  terres- 
trial meridian  of  the  observer's  standpoint. 

The  intersections  of  the  celestial  meridian  and  horizon  are 
the  North  and  South  Points. 

(6)  The  Prime  Vertical  is   that  vertical   circle  which  is 
perpendicular  to  the  celestial  meridian. 

The  points  of  intersection  of  the  prime  vertical  and  the 
horizon  are  the  East  and  West  Points. 

(7)  The  Ecliptic  is  that  great  circle  of  the  celestial  sphere 
in  which  the  sun  appears  to  move. 

The  intersections  of  the  equinoctial  and  ecliptic  are  called 
the  Vernal  and  Autumnal  Equinoxes.  The  sun  appears  to 
be  at  these  points  on  about  March  21,  and  September  21, 
respectively. 

The  angle  between  the  equinoctial  and  ecliptic  is  called 
the  Obliquity  of  the  Ecliptic.  It  is  denoted  by  the  letter  e. 
e  =  23°  27',  nearly. 

(8)  The  hour  circle  passing  through  the  equinoxes  is  called 
the  Colure. 

(9)  Latitude   Circles   are    the    great   circles   which   pass 
through  the  poles  of  the  ecliptic. 


82  SPHERICAL   TRIGONOMETRY. 

(10)  That  latitude  circle  which  passes  through  the  equi- 
noxes we  shall  call  the  Prime  Latitude  Circle. 

b.    Spherical  Coordinates. 

1.  Definition.     The   Spherical  Coordinates  of   a  heavenly 
body  are  its  angular  distances  from  two  fixed,  mutually  per- 
pendicular, great  circles  of  reference. 

For  convenience  the  great  circle  arcs  which  measure  the 
angular  distances,  are  used. 

2.  Systems  of  Spherical  Coordinates. 

(1)  Using  the  horizon  and  meridian  as  reference  circles, 
we  get  Altitude  and  Azimuth,  respectively  ;    altitude  taken 
from  the  horizon  from  0°  to  90°,  and  azimuth  taken  from  the 
south  point  measured  by  way  of  the  west  from  0°  to  360°. 
The  complement  of  the  altitude  is  called  the  Zenith  Distance. 

(2)  Using  the  equinoctial  and  meridian  we  get  Declination 
and  Hour  Angle,  respectively  ;    declination  taken  from  the 
equinoctial  north  or  south  from  0°  to  90°  or  —  90°,  and  hour 
angle  taken  from  the  meridian  measured  by  way  of  the  west 
from  0°  at  noon  to  360°,  or  24  hrs.     The  complement  of  the 
declination  is  called  the  Polar  Distance. 

(3)  Using  the  equinoctial  and  colure  we  get  Declination 
and  Eight  Ascension,   respectively;    right   ascension   taken 
from  the  vernal  equinox  by  way  of  the  east  from  0°  to  360°, 
or  24  hrs. 

(4)  Using  the  ecliptic  and  prime  latitude  circle  we  get 
Celestial   Latitude    and   Celestial    Longitude,    respectively ; 
celestial  latitude  taken  from  the  ecliptic  from  0°  to  90°  or 
—  90°,  and  celestial  longitude  taken  from  the  vernal  equinox 
by  way  of  the  east  from  0°  to  360°. 

It  is  important  not  to  confound  celestial  latitude  and  longi- 
tude with  terrestrial. 

Pigs.  41  and  42  represent  the  celestial  sphere,  with  its 
center  at  0. 


SPHERICAL  CO  ORDINA  TES. 


83 


In  Fig.  41,  NWS  represents  the  horizon  ;  EQR  the  celestial  equator; 
Z,  Na,  and  Pn ,  Ps,  their  respective  poles;  Ve,  the  vernal  equinox ;  and 
H,  a  heavenly  body. 

AH     =  altitude        denoted  by  h. 

SWA  =  azimuth       «  a. 

DH     =  declination       "  5. 

Angle  RPnH  =  Arc  RWD  =  hour  angle "  t. 

VeD    =  right  ascension "  a. 

RZ     =  NPn  =  terrestrial  latitude  of  observer's  standpoint        "  0. 
ZH     =  90°  —  h  =  zenith  distance. 
pn  H    =  90°  —  5  =  polar  distance. 

The  triangle  PnHZ  is  called  the  Astronomical  Triangle  ;  the  angle  at 
H,  the  Position  Angle.  The  student  should  carefully  note  the  parts  of 
this  triangle. 


84 


SPHERICAL   TRIGONOMETRY. 
G 


FIG.  42. 

In  Fig.  42,  QVeR,  represents  the  equinoctial ;  EVeC,  the  ecliptic  ;  Pn, 
Ps,  and  F,  G,  their  respective  poles;  Ve,  the  vernal  equinox;  and  H,  a 
heavenly  body. 

DH  =  declination. 
VeD  =  right  ascension. 
LH  =  celestial  latitude,  denoted  by  £. 
VeL  =  celestial  longitude,         "         X. 
CR  =  PnG  =  obliquity  of  the  ecliptic. 

The  student  should  carefully  note  the  parts  of  the  triangle  PnHG. 

•> 

n.  Examples.    X. 

1.  a  =  116°,  b  =  16°,  C  =  90°. 

2.  A  =  60°  47',  B  =  57°  16',  c  =  90°. 

3.  a  =  28°,  b  =  28°,  B  =  79°. 


EXAMPLES.     X.  85 


4.  a  =  33°  40',  A  =  43°  21',  C  =  90°. 

5.  A  =  105°  53',  a  =  104°  54',  c  =  90°. 

6.  a  =  61°,  B  =  123°  40',  C  =  90°. 

7.  A  =  139°,  b  =  143°,  c  =  90°. 

8.  a  =  69°  18',  c  =  84°  27',  C  =  90°. 

9.  A  =  70°  12',  C  =  106°  25',  c  -  90°. 

10.  c  =  69°  25',  A  =  54°  54',  C  =  90°. 

11.  C  =  134°,  a  =  143°,  c  =  90°. 

12.  A  =  46°  59',  B  =  57°  59',  C  =  90°. 

13.  a  =  174°  12',  b  =  94°  08',  c  =  90°. 

For  a  right  spherical  triangle  prove,  when  C  =.90°  : 

14.  tan2  -  =  tan  •$•  (a  +  c)  tan  £  (c  —  a). 

15.  tan2  (45°  -  $  A)  =  tan  $  (c  -  a)  ctn  $  (c  +  a). 

16.  tanH  B  =  sin  (c  —  a)  esc  (c  +  a). 

17.  tan2  (45°  —  \  B)  =  tan  £  (A  —  a)  tan  -J-  (A  +  a). 

18.  tan2  (45°  —\  b)  =  sin  (A  -  a)  esc  (A  +  a). 

19.  tan2  (45°  -  i  c)  =  tan  i  (A  -  a)  ctn  £  (A  +  a). 

tan  j  (90°+  B-A) 


21.  ctn2  i  b  =  tan  |  (90°  +  A  -  B)  ctn  $  (B  +  A  -  90°). 

22.  ctnHc  =  —  cos(B  —  A)sec(B  +  A): 

23-31.    Obtain  formulas  for  the  quadrantal  triangle  corresponding  to 
formulas  of  problems  14-22. 

32.  A  spherical  square  is  a  spherical  quadrilateral  which  has  equal 
sides  and  equal  angles.     The  diagonals  divide  it  into  four  equal  spherical 
right  triangles.     Given  a  side  a  of  the  spherical  square,  find  an  angle  A. 

33.  A  line  makes  with  a  plane  an  angle  a  ;    through  the  foot  of  the 
line  and  in  the  plane  a  line  is  drawn  making  an  angle  ft  with  the  projec- 
tion of  the  first  line  on  the  plane  ;  find  the  angle  7  made  by  the  second 
line  with  the  first. 

34.  Given  the  number  of  sides  of  a  regular  spherical  polygon  equal  to 
n  and  each  angle  equal  to  A  ;   find  a  side  of  the  polygon  and  the  polar 
radii  of  the  inscribed  and  circumscribed  circles. 

35.  Find  the  angles  between  the  adjacent  faces  of  each  of  the  five 
regular  polyhedrons. 

36.  The  base  of  a  regular  pyramid  is  an  octagon  ;   each  angle  at  the 
vertex  of  the  pyramid  is  30°;   find  the  angle  a  between  two  adjacent 
lateral  faces,  and  the  angle  ft  between  any  lateral  face  and  the  base. 

37.  From  the  longitude  of  the  sun  (A)  and  the  obliquity  of  the  ecliptic 
(e)  find  the  sun's  right  ascension  (a)  and  declination  (8). 


86  SPHERICAL  TRIGONOMETRY. 

38.  From  the  latitude  (0)  of  a  place  on  the  earth's  surface  and  the 
declination  (5)  of  the  sun  on  a  given  day,  find  the  times  and  places  of  the 
rising  and  setting  of  the  sun  and  its  distance  from  the  zenith  at  noon. 

When  are  days  and  nights  equal  ? 
(Neglect  the  effect  of  refraction.) 

39.  When  does  the  solution  of  Ex.  38  become  impossible?      When 
indeterminate  ?      And  what  follows  for  places  so  situated  on  the  earth's 
surface  as  to  give  these  results  ? 

40.  When  and  where  does  the  sun  rise  in  Oberlin  (latitude  41°  17')  on 
the  longest  day,,  June  21  (5  =  +23°  27")  ?     When  and  where  on  the 
shortest  day,  Dec.  21  (5  =  -  23°  27')  ? 

41.  At  a  place  on  the  earth's  surface  whose  latitude  is  0,  a  rod  OA, 
pointed  toward  the  north  pole,  makes  with  a  horizontal  plane  an  angle, 
BOA  =  0;   let  OC  be  the  position  of  the  shadow  of  the  rod  at  a  given 
time  of  the  day  ;   find  the  angle  BOC. 

What  does  the  result  become  for  0  =  41°  17',  t  =  2  hrs.  =  30°? 

42.  Through  0  (Ex.  41)  pass  a  plane  perpendicular  to  OB  ;   extend 
the  rod  through  this  vertical  plane  so  that  its  shadow  may  fall  upon  it ; 
find  the  angle  S  which  the  shadow  makes  with  the  projection  of  the  rod 
on  the  plane  at  a  given  time  t.     Find  S  for  0  =  41°  17',  t  =  3  hrs.  =  45°. 

43.  At  what  time  of  a  given  day  (5  given)  at  a  given  place  (lat.  =  0) 
is  the  sun  exactly  east  or  west  ? 

44.  Find  the  altitude  (h)  and  the  azimuth  (a)  of  the  sun  at  a  given 
place  and  on  a  given  day  at  6  A.M. 

45.  Find  the  shortest  distance  (i.e.,  length  of  great  circle  arc)  from 
Oberlin,  long.  82°  14',  to  New  Haven,  long.  72°  55',  the  latitude  of  each 
place  being  41°  IT, 


CHAPTER   III. 
A.   Formulas  for  Spherical  Oblique  Triangles. 

a.   Formulas  for  Functions  of  £  A,  £  B,  £  C. 

§  91.   From  the  formulas  of  §  77, 

cos  a  —  cos  b  cos  c 

sin  b  sin  c 

^/1-f  cos  A /cos  a  —  (cos  b  cos  c  —  sin  b  sin  c) 

2  *  2  sin  b  sin  c  c  f 


Vcos  a  —cos  (b  -f-  c)  _      /sin  ^  (a  -J-  b  +  c)  sin  -j-  (b  -(-  c  —  a) 
2  sin  b  sin  c  ^  sin  b  sin  c  §  59 


Let 


/sm  s  sin  (s  —  a) 

and  cos  4-  A  =  \/ : — ; — » 

^       sm  b  sm  c 


/sin  s  sin  (s  —  b) 
Similarly,      cos  £  B  =  V—         -A L 

*         sin  r  sin  a 


/sin  s  sm  (s  —  c) 

~  \    ,     ' 

sm  a  sm  b 


§92. 


sin  b  sin  c 

§§  58  and  59 


Q.     .,     ,  -ID          /sin  (s  —  c)  sin  (s  —  a) 

Similarly,     sin  -J-  B  =  \  —  —  • 


sm  c  sm  a 


.n  Jsin(s-a)sin(s-b) 

~ 


siri  a  sin  b 


88  SPHERICAL   TRIGONOMETRY. 

§  93.    From  the  formulas  of  §§  91  and  92, 


. 
sin  s  sin  (s  —  a) 


sin  s  sin  (s  —  b) 


sin  s  sin  (s  —  c) 

The  student  may  prove  that  in  each  formula  of  the  last 
section  the  positive  square  root  is  to  be  taken,  and  that  the 
number  under  the  radical  sign  in  each  case  is  positive. 

§  94.  From  the  formulas  of  §  81,  or  from  the  formulas  of 
the  last  three  sections,  by  using  the  properties  of  polar 
triangles,  the  student  may  prove  the  following  formulas  : 

/cos  (S  —  B)  cos  (S  —  C) 
cos     a          -  —  —  - 


=  \/ 
* 

/ 

=  \ 
* 


. 
sin  B  sin  C 


/  —  cos  S  cos  (S  —  A) 
sm     a  =     -  .    D    .v     -  L 
sin  B  sin  C 


j  /    —  cos  S  cos  (S  —  A) 

=  ^cos(S-B)cos(S-C)' 

Where  S  =  A"["^  +  C  • 

The  student  may  write  the  corresponding  formulas  for 
b  and  c,  and  prove  that  in  each  formula  the  positive  square 
root  is  to  be  taken,  and  that  the  number  under  the  radical 
sign  in  each  case  is  positive. 

b.    Gauss's  Equations. 
§  95.    From  §§54  and  55 

cos  -j-  (A  ±.  B)  =  cos  %  A  cos  J  B  rp  sin  -J-  A  sin  £  B. 
Substituting  for 

cos  \  A,  cos  \  B,  sin  J  A,  and  sin  £  B, 
their  values  given  in  §§  91  and  92  : 


SPHERICAL   OBLIQUE   TRIANGLES.  89 


cos  |(A  ±  B)  = 

7 


sin  c   *  sin  a  sin  b 


—  s*n  (s  ~  c)     /sin  (s  —  a)  sin  (s  —  b) 
sin  c        ^  sin  a  sin  b 

sin  s  rp  sin  (s  —  c)    .    C 
=  -  — ^sm  — • 

sin  c  2 

Taking  the  upper  sign, 

c\  _     .    /a  +  b      c\ 

cos  J  (A  +  B)  = ^— ?f — ^-  sin  ^ 

2  sin  -  cos  - 

2  cos  i  (a  +  b)  sin  -        p 

____2sinC  §59 

2  sin  -  cos  - 


c 

cos- 


Taking  the  lower  sign, 

.  sin  s  +  sin  (s  —  c)    .     C 

cosi(A-B)=-      --      ^sm- 


sin  — 
Similarly  from 

sin  %  (A  ±  B)  =  sin  %  A  cos  -J  B  ±  cos  |-  A  sin  -J  B. 
Substituting  as  above, 

sin  (s  —  b)  ±  sin  (s  —  a)        C 
sin  i  (A  ±  B)  =  -  J .        — ^ — — z  cos  - 

sin  c  2 

,   a-b\  /c      a-b 


2  sin  -  cos  - 


90  SPHERICAL   TRIGONOMETRY. 


.                          cos  J  (a  —  b)        C 
.'.  sin  -J-  (A+  B)  =—        * ^cos  —  > 

cos  I 

sin  %  (a  —  b)        C 
and  sin  £  (A  —  B)  =  -  — *  cos  —  • 


These  formulas  may  be  put  in  a  form  in  which  they  may 
be  more  easily  remembered.     They  can  be  written  as  follows  : 

.          .    .   Dx                 ,        ,  Here    observe  :     (1) 

sm-HA+B)      cos  -J-  (a  —  b)  ...  ,,                   .    _V 

1.   v      ' — z  =  —              — z  •  All  the  angles  are  half 

cos  *  C                 cos  *  c  , 

.     ,  /A       DX         •     i,        .v  angles.     (2)  In  the  left 
_     sin  i  (A  —  B)       sin  %  (a  —  b)  3  ,       v  ' 

2.   v     _ — ^  = =-r-j £  •  members  of  the  equa- 

cos  i  C  sin  4-  c  , .  i          *    iW 

.  tions  we  have  A,  B,  C  ; 

_     cos  |  (A  +  B)      cos  £  (a  -f  b)  . 

3.    •     i  /^ — '  — ^ di  ln  the  right  members 

sin  j-  C  cos  i  c  ® 

i  /  A       D\         •     i  /     i   u\  a,  b,  c.    (3)  In  each  left 
cos  4-  (A  —  B)      sin  4-  (a  +  b)  n      v  ' 

4.    .  v  .  _ — L  = .  \   — -'  member     are     co-func- 

sm  ^  C  sm  £  c  , .          .  .  .  ^   J_. 

tions,  in  the  right,  the 

same  functions.  (4)  In  each  equation  to  a  sign  -f~  in  the 
numerator  of  one  member  corresponds  a  cosine  in  the  numer- 
ator of  the  other  member.  And  in  the  same  way  to  a  sign  — 
corresponds  a  sine. 

c.  Napier's  Analogies. 

§  96.   If  we  divide  member  by  member,  1  of  the  Gauss 
equations  by  3,  2  by  4,  4  by  3,  and  2  by  1,  we  obtain : 

i  /A    i   D\  i  /  (1)   Here  the   parts 

tan  4-  (A  +  B)      cos  4-  (a  —  b)  v  ' ,, 

1.  -       1     ,  i — *  = ¥/ — r~R  '     enter  nve  at  a  time,  and 

ctn  -J-  C            cos  ^  (a  -f  b)  .     ^. 

.  ,  A       DN         •     f  /         ,  x  m  the  respective  mein- 

_     tan  -J-  (A  —  B)       sm  i  (a  —  b)  ,                            .     _     _ 

2.   1     -,  ^ — L  =  - — H — rix  *  bers  we  nave   A>  B,  C 

ctn  ^  C  sin  }  (a  +  b) 

i  /     i   .  N  i  /A       o\       an« a?  b  5  or  a?  D,  c,  and 

tani(a+b)=cos^A-B   .         g_  ' 

tanjc  COSHA  +  B        enter  tVe  left  members 

tan  4-  (a  —  b)        sm  4-  (A  —  B) 

4.    — -^T L  =- — ;  ;.    ,      '•     we  have  there  the  co- 

tan  -J-  c  sm  -J-  (A  -f  B) 

functions,  tangent  and 


SPHERICAL   OBLIQUE   TRIANGLES.  91 

cotangent,  but  the  same  functions,  tangents,  when  a,  b,  c, 
enter  the  left  members.  In  the  right  members  are  the  same 
functions,  sines  or  cosines.  (3)  The  angles  are  half  angles. 

(4)  In  the  right  members  are  sines  or  cosines  according  as 
the  signs  —  or  +  are  in  the  numerators  of  the  left  members. 

(5)  In  the  right  members  the  angles  are  differences  in  the 
numerators,  sums  in  the  denominators. 

d.  1'Huilier's  Formula. 

8 97.   If  in  1  and  3,  §  95,  A+^+-C  —  ^  be  represented 

2i  2i 

byE, 

/C     c\ 

cos  (  —  —  L                ,  ,        , . 
\2         I      cos  j  (a  —  b) 
we  get  — £—  *-£ L> 

cos  -  cos  - 


sin(§-E)      cosHa  +  b) 

C 

sm- 

cos| 

and 

By  division  and  composition  the  first  of  these  gives 

cos  f  —  —  E  J  —  cos  —      cos  -J-  (a  —  b)  —  cos  - 

(C         \  C=  c       '>sou, 

-  —  E  J  +  cos  -     cos  %  (a  —  b)  +  cos  - 

—  tan  £  (C  —  E)  tan  £(—£)=  —  tan  |  (s  —  b)  tan  £  [— (s— a)], 
or,    tan£(C  —  E)  tan£  E        —      tanj-  (s  —  b)  tan^-(s  —  a). 

In  a  similar  manner  the  second  equation  gives, 

ctn  £  (C  —  E)  tan  $  E  =  tan  £  s  tan  i  (s  —  c)  ; 

multiplying  these  equations  together, "term  by  term,  we  have, 

tan2!-  E  =  tan  %  s  tan  £  (s  —  a)  tan  £  (s  —  b)  tan  •$•  (s  —  c). 


92 


SPHERICAL   TRIGONOMETRY. 


Formula  for  the  Polar  Radius  of  a  Circumscribed  Circle 
of  a  Spherical  Triangle.  , 

§  98.  The  construction  is  anal- 
ogous to  that  for  the  similar 
problem  in  the  case  of  a  plane 
triangle.  P,  the  pole  of  the 
required  circle  is  the  intersec- 
tion of  perpendicular  great  circle 
arcs  erected  to  the  sides  at  their 
middle  points. 

the  required  radius. 


O A  n C D  C f* 

In  the  right  spherical  triangle  LPC,  by  §§  86  and  94, 
tan- 


tan  R  = 


—  cos  S 


cos  (S  —  A) 
= =  tan  -J  a  tan  -J  b  tan  -J  c,  by 

~~  COS  o 


cos  (S  —  A)  cos  (S  —  B)  cos  (S  -  C) 
94. 


/.  Formula  for  the  Polar  Radius  of  the  Inscribed  Circle 
of  a  Spherical  Triangle. 

§  99.  Bisect  the  angles  of  the 
triangle  by  great  circle  arcs. 
From  the  point  of  intersection  of 
the  angle  bisectors  draw  perpen- 
dicular arcs  of  great  circles  to  the 
sides  of  the  triangle.  Since  L,  M, 
and  N  are  points  of  tan  gen  cy, 
BN  =  BL,  AN^AM,  CM  =  CL, 
a'+b'  =  c,  b'+c'  =  a,  c'-f  a'  =  b, 
adding  and  dividing  by  2,  FIG.  44. 


DEDUCTION  OF  PLANE   TRIGONOMETRY. 


93 


a-   I    b'   I   c'-a+b  +  C-c 

2 

.•.a'  =  s-(b'+  c')  =  (s-a). 
A       tan  r 


tan  r 


In  the  right  triangle  AN  P.  tan  —  =  — : .  =  -; — 

2      sm  a'      sin  (s  —  a) 

.'.  tan  r  =  tan  —  sin  (s  —  a) 

L 


-V" 


sin  (s  — a)  sin  (s  —  b)  sin  (s  —  c) 


sin  s 
—  sin  s  tan  -J  A  tan  -J-  B  tan  -J-  C. 


§86 


§  93 


B.    Plane  Trigonometry  a  Special  Case  of  Spherical 
Trigonometry. 

§  1OO.  Let  ABC  be  any  spherical  triangle  of  which  the 
sides  a,  b,  and  c  are  measured  in  some  linear  unit.  Let  r  be 
the  radius  of  the  sphere,  and 

a,  /?,  and  y  the  angles  at  the 
center    corresponding    to    the 
arcs  a,  b,  and  c,  respectively. 

Then   «  =  2      p  =  ±      y  =  ? 
by  §  8,  3. 

If  now  r  increases  while  a, 

b,  and  c  remain  fixed  in  length, 
a,  (3)  and  y  will  diminish  ;  and 
if  r  increases  indefinitely,  a,  /?, 
and  y  will  approach  zero  as  a 
limit.     And  the  limit  as  r  in- 
creases indefinitely  of 

lim          sin  a         lim 


FIG.  45. 


lim    sin  a 


r  sin  a  =  _  r  a 

a  =  0  a 


sm  a  _ 
a  =  0        a  a  =  0a 


=  a. 


Similarly  as  r  increases  indefinitely  lim  r  sin  ^  =  ?> 


94  SPHERICAL   TRIGONOMETRY. 

We  have  by  §  77   cos  a  =  cos  /?  cos  y  -\-  sin  (3  sin  y  cos  A. 
cos  a  =  1—2  sin2  -  ,  by   §  58.     Substituting   this   value   for 

cos  a  and  similar  values  for  cos  (3  and  cos  y  in  the  above 
equation  and  reducing  we  have  : 

in2  1  =  sin2  ^  -f  sin2  1  —  2  sin2  1  sin2  1  —  -J-  sin  ft  sin  y  cos  A. 

Multiplying  this  equation  through  by  r2  and  taking  the  limits 
of  both  members  as  r  increases  indefinitely  we  have  : 


sn 


(r  sin  \  J=  (r  sin  |J+  (r  sin  *J-  ?  (r  sin  f)'(r  sin  * 

—  -J-  (r  sin  /3)  (r  sin  y)  cos  A. 


an  equation  which  with  two  others  similar  to  it,  and  obtained 
in  the  same  way,  may  be  considered  the  fundamental  equa- 
tions of  plane  trigonometry.  This  has  led  us  to  the  broadest 
generalization  of  which  we  are  capable  at  this  stage,  but  the 
student  will  learn  subsequently  that  all  trigonometry  of  the 
sphere  and  plane  is  wrapped  up  in  one  quaternion  equation, 
an  equation  which  in  the  quaternion  calculus  is  very  elemen- 
tary. This  equation  is  : 

S  (VyoVajS)  =  a2S/?y  —  SyaSa/?, 

and  being  interpreted  with  reference  to  a  sphere  it  gives  at 
once  the  three  equations  of  §  77,  which  are  the  fundamental 
equations  of  spherical  trigonometry,  which  includes  plane 
trigonometry  as  a  special  case. 

C.  Spherical  Triangles. 

(f.   Cases. 

§  1O1.    With  any  three  parts  given,  a  spherical  triangle 
may  be  solved,  and  the  equations  of  §  77  theoretically  deter- 


SPHERICAL    TRIANGLES.  95 

mine  the  solution.     In  practice,  however,  derived  equations 
adapted  to  logarithmic   computation  are  used.      The  three 

given  parts  may  be  selected  in    '   '    =  20  ways,  as  follows  : 

ABC  ACa  Aarc  BCc  Cab 

ABa  ACb  Abe  Bab  Cac 

ABb  ACc  BCa  Bac  Cbc 

ABc  Aab  BCb  Bbc  abc 

Of  these,  there  are  only  six  distinct  cases. 
Excluding  repetitions,  we  have  : 

1.  abc,    three  sides. 

2.  ABC,  three  angles. 

3.  abC,   two  sides  and  included  angle. 

4.  ABc,  two  angles  and  included  side. 

5.  abA,    two  sides  and  angle  opposite  one. 

6.  ABa,  two  angles  and  side  opposite  one. 

The  number  of  these  cases  may  again  be  reduced  by  one 
half,  by  the  use  of  the  polar  triangle,  since,  for  example, 
cases  2,  4,  and  6  could  be  made  to  depend  upon  cases  1,  3, 
and  5,  respectively.  But  it  is  practically  better  to  know 
how  to  treat  six  cases. 

b.   Solutions. 

Case  1.  Given  a,  b,  and  c.  In  order  that  the  triangle 
should  be  possible,  the  following  inequalities  must  be  satisfied  : 


a  +  b>c>0  7r>a>0 

b  +  c>a>0  7r>b>0 

c  +  a>b>0  TT>C  >0 

If  these  inequalities  are  satisfied,  the  solution  is  determinate 
in  a  single  way,  and  the  angles  are  given  by  the  formulas  : 


96  SPHERICAL   TRIGONOMETRY. 

A  tan  r  B  tan  r  C  tan  r 


2       sin  (s  —  a)  2       sin  (s  —  b)         '2       sin  (s  —  c) 

/sin  (s  —  a)  sin  (s  —  b)  sin  (s  —  c") 

where      tan  r  =  +  \  —  -A —  — z  •      §  99 

^  sin  s 

If  only  a  single  angle  is  desired,  it  may  be  computed  from 
one  of  the  formulas, 

A  /sin  (s  —  b)  sin  (s  —  c) 

tan  -  =  +  \l ^ /   .    v — — A  etc.  §  93 

2  *       sm  s  sin  (s  —  a) 

As  a  check,  the  formulas 

sin  a sin  b sin  c 

sin  A      sin  B      sin  C 
may  be  used. 

Case  2.    Given  A,  B,  and  C.     If  the  following  inequalities, 


27T>7T+A>B+C,  7T>A>0, 

27T>7T+B>C+A,  7T>B>0, 

27T>7T+C>A+B,  7T>C>0, 

* 

are  satisfied,  the  three  sides  may  be  obtained  determinately 
in  a  single  way,  from  the  formulas  : 

tan  -  =  tan  R  cos  (S  —  A), 

tan  -  =  tan  R  cos  (S  —  B), 

2i 

tan  —  =  tan  R  cos  (S  —  C), 


V_  pr)c   ^ 
7F K 7^ 5^ 7^ 7^'  §  98 
cos  (S  —  A)  cos  (S  —  B)  cos  (S  —  C) 

The  formulas  of  §  79  may  be  used  as  a  check.  If  only  a 
single  side  is  desired,  one  of  the  formulas  of  §  94  may  be 
used. 


SPHERICAL   TRIANGLES.  97 

Vase  8.    Given  a,  b,  and  C.     If  the  inequalities, 

7r>a>0,  TT  >  b  >0,  TT  >  C>  0, 

are  satisfied,  we  may  solve  the  triangle,  which  will  be  deter- 
minate in  a  single  way,  as  follows  :  A  and  B  may  be  found 
by  using  the  formulas  : 

a-b 
A+B      COS~2-        C 


COS 


2 
a-b  §96 


A_B 

~ 


A+  B 
These  give  us          —  -  —  =  m, 

2  A  =  m  -f  n, 

A        D  whence    _ 

A  —  B  _  B  =  m  —  n. 

~~2~ 
c  may  be  found  from  the  formula  of  §  96  : 

A  +  B 
c      S1"^T-         a-b 


As  ?,  check  use  §  79. 

If  the  side  c  alone  is  desired,  it  may  be  found  from  the 
formula,  §  77, 

cos  c  =  cos  a  cos  b  4-  sin  a  sin  b  cos  C, 

which  may  be  adapted  to  logarithmic  computation  by  the  aid 
of  an  auxiliary  angle,  as  follows  : 

cos  c  =  cos  b  (cos  a  +  sin  a  tan  b  cos  C). 

If  tan  b  cos  C  is  put  equal  to  tan  x  =  --  >  we  have  the 

cos  x 

two  formulas, 

cos  b  cos  (a  —  x)  _. 

cos  c  =  -  *  -  *  >   and  x  =  tan"1  (tan  b  cos  C), 
cos  x 


SPHERICAL   TRIG  ONOME  TR  Y. 


or  a  single  formula, 

_  cos  b  cos  [a  —  tan"1  (tan  b  cos  C)] 

cos  tan"1  (tan  b  cos  C) 

In  the  same  manner,  A  and  B  can  be  computed  singly. 
§  80,  we  have 

ctn  a  sin  b  =  cos  b  cos  C  -f-  sin  C  ctn  A, 
ctn  b  sin  a  =  cos  a  cos  C  +  sin  C  ctn  B, 
or  sin  C  ctn  A  =  ctn  a  (sin  b  —  cos  b  tan  a  cos  C), 

sin  C  ctn  B  =  ctn  b  (sin  a  —  cos  a  tan  b  cos  C). 
If  we  put  tan  b  cos  C  —  tan  x, 

tan  a  cos  C  =  tan  y, 
these  formulas  become 

_  ctn  a      sin  (b  —  y)  _  ctn  C  sin  (b  —  y) 
~  sin  C  cos  y  sin  y 


By 


Ctn  D  — 


or 


ctn  b      sin  (a  —  x)  _  ctn  C  sin  (a  —  x) 
sin  C  cos  x  sin  x 

ctn  C  sin  [~b  —  tan-1  (tan  a  cos  C)l 

ctn  A  — - * ) 

sin  tan-1  (tan  a  cos  C) 

ctn  C  sin  [a  —  tan"1  (tan  b  cos  C)] 

sin  tan-1  (tan  b  cos  C) 

The  employment  of  the  auxil- 
iary angles  x  and  y  is  equiva- 
lent to  decomposing  the  triangle 

ABC 

into  two   right   triangles, 
abc 

For  if  AD  is  an  arc  perpendicular 
to  BC  from  A, 

tanx'  =  tanbcosC,  §  86,  4 
whence  x  =  x'. 


FIG.  46. 


tan  AD  = 


Also,  cps 

sin  x 
ctn  C 


COS  X 


§    86,1 
§  86,  6 


SPHERICAL   TRIANGLES.  99 

And  in  the  triangle  ADB  we  have 

cos  b  cos  (a  —  x) 

cos  c  =  cos  AD  cos  (a  —  x)  =  -  »  -  L> 

cos  x 

_       sin  (a  —  x)       ctn  C  sin  (a  —  x) 
and         ctn  b  =  —  —  -  —  .  ^  '  =  -  :  — 
tan  AD  sin  x 

In  the  same  way.  if  B  E  is  a  perpendicular  arc  from  B  to 
AC,  y  =  y',  and  the  angle  A  may  be  found  similarly  as  the 
angle  B. 

Case  4.  Given  A,  B,  and  c.  If  the  inequalities  TT  >  A  >0, 
TT  >  B  >0,  7r>c>0,  are  satisfied,  the  triangle  'is  possible, 
and  determinate  in  a  single  way.  We  may  find  a,  and  b  by 
the  following  formulas  : 

A-B 
a+b      COS-T~         c 


cos 


sin 


A-B 


a  —  b  2  c 

tan  —^—  A  ,    B  tan  -  •     These  give 


~2~~P>  a  =  p  +  q, 

whence 

a  —  b  _  b  =  p  —  q. 

~T~  ~q' 

C  may  be  found  from  the  formula  of  §  96, 

a+  b 
c      sin  -y-         A  _  B 

ctn  —  = tan  -      —  •     Use  §  79  as  a  check. 

£i  a —  b  Z 


sin 


2 


If  only  the  angle  C  is  required,  it  may  be  obtained  by  the 
formula,  §  81, 


100  SPHERICAL    TRIGONOMETRY. 

cos  C  =  —  cos  A  cos  B  -f-  sin  A  sin  B  cos  c 
=  cos  B  ( —  cos  A  -f-  sin  A  tan  B  cos  c) 
_  cos  B  sin  (A  —  u) 

sin  u 
_  cos  B  sin  [A  —  ctn-1  (tan  B  cos  c)] 

sin  ctn-1  (tan  B  cos  c) 
if  ctn  u  =  tan  B  cos  c. 

Similarly  the  sides  a  and  b  may  be  computed,  if  in  the 
formulas, 

ctn  a  sin  c  =  cos  c  cos  B  -j-  sin  B  ctn  A, 
ctn  b  sin  c  =  cos  c  cos  A  +  sin  A  ctn  B, 
we  put  ctn  u  =  tan  B  cos  c, 

ctn  v  =  tan  A  cos  c, 

ctn  c  cos  FB  —  ctn-1  (tan  A  cos  c)] 

and  obtain,       ctn  a  = u r- ; ; : -**  > 

cos  ctn"1  (tan  A  cos  c) 

ctn  c  cos  [A  —  ctn-1  (tan  B  cos  c)] 

cos  ctii"1  (tan  B  cos  c) 

Similarly  as  in  Case  3,  it  may  be  shown  that  the  use  of  the 
auxiliary  angles   u  and  v,  is  equivalent  to  decomposing  the 

triangle          '   into   two   right   triangles.     The   student  may 

8.  DC 

that  angle  u  =  angle  BAD,  . 
prove       ,  &,     A  _    '  in  Fig.  46. 

and   angle  v  =  angle  ADD, 

Case  5.    Given  a,  b,  and  A. 

(1)  If  the  inequalities,  7r>a>0,  TT  >  b  >  0,  7r>A>0, 
are  satisfied,  there  will  always  be  one  solution,  when  b  differs 

more  than  a  from  —  • 

(2)  Unless    the     inequalities     : —      -  [=  sin  B]  <  1, 

sin  a 

TT  >  .  >  — ,  or  else  —  >     >  0,  and  TT  >  b  >  0,  are  satisfied 

r\  £  +J  r\ 

when  a  differs  more  than  b  from  — ,  there  will  be  no  solu- 


SPHERICAL    TRIANGLES.  101 

tion  ;  and  when  they  are  satisfied,  and  a  and  b  thus  related, 
there  will  always,  necessarily,  be  two  solutions. 

These  statements  may  be  proved  as  follows  :  It  is  evident 
that  there  can  be  no  solution  under  (2),  unless  the  inequali- 
ties, TT  >  ^  >  ^ ,  or  else  ^  >  *  >  0,  are  satisfied  under  the 
conditions.  For,  in  any  spherical  triangle  with  real  parts,  if 
one  side  differs  more  than  another  side  from  — ,  its  opposite 
angle  is  of  the  same  quadrant  with  it. 

This  appears  from  the  fact  that  the  parts  of  the  triangle 

cos  a  —  cos  b  cos  c       __        , 

must  satisfy  the  relation  cos  A  = : — ; — : >  §77,  and 

sin  b  sin  c 

if  a  differs  more  than  b  from  —i  cos  a  >  cos  b,  numerically, 

and    if    c    is    real,    cos  c  <  1,    numerically,    and,    therefore, 
cos  a  >  cos  b  cos  c,  numerically,  and  hence,  cos  A  and  cos  a 
have  the  same  sign,  and  A  and  a  are  of  the  same  quadrant. 
To  prove  the  remaining  statements,  we  consider  the  equa- 
cos  a  —  cos  b  cos  c 


tion  sin  c  = 


sin  b  cos  A 


cos  a  cos  b  , , 

where  we  let   — —     r  =  m,    — — r—  — r  =  n,    and  get  the 

sin  b  cos  A  sm  b  cos  A 

following  equation  in  sin  c  : 

(1  +  n2)  sin2c  —  2  m  sin  c  +  m2  —  n2  =  0. 

>2_  n2 


The  product  of  the  roots  is  — 2      "  • 
The  roots  are  : 

sin  G!  —  •  2          (m  +  n  (n2  +  1  —  m2)^), 

sin  c2  =  -^-7-  --  (m  —  n  (n2  +  1  —  m2)1). 


IS 


102  SPHERICAL   TRIGONOMETRY. 

According  as  4      I  differs  more  than  \      i  from  —  >  that  i 

according  as  we  have  •{,<>;}•,  n'2  ^  m2,  and  in  (2)  m  >  0. 

I  (*)  J          ^ 

According  as  we  have  •!  xo\  f »  *^e  Pr°duct  °f  the  roots, 

m2— n2    .  „         ("negative")  . 

— 0  .  „  >  is  therefore  s  >  ;    and  its  form  shows  that 

ir  +  1  L  positive  J 

the  factors,   i.e.,  the  roots,   must  be  real,   and   that,   there- 
fore  n2-fl  —  m2>0;    and  in  \  ;  ^  V  that  the  roots  must  be 

iw.J 

f  positive  and  negative  "]  . 

4  fT,          ...  ..      h  •      And  m  (2)  both  roots 

[both  positive  or  both  negative j 

must  be  positive,  since,  as  m  >  0,  a  single  root  at  least  must 
be  positive. 

Again,        2  m2n2  (n2  +  1  -  m2)  -<  m4n4  +  (n2  + 1  —  m2)2, 
and 

4  m2n2  (n2  + 1  —  m2)  — <  m4n4+ 2  m2n2  (n2+ 1  —  m2) 

-h(n2+l-m2)2. 

Taking  positive  values  of  m,  n,  and  n2-[-l—  m2, 
2  mn  (n2  +  1  -  m2)*-<  n2  +  1  -  m2+  m2n2, 
.-.2mn(n2  +  l  — m2)^— <n4  +  2n2+l  — m2— n2(n2  +  l  — m2), 
. ' .  m2  +  n2  (n2  + 1  —  m2)  +  2  mn  (n2  +  1  —  m2)*  — <  (n2  +  I)2, 
.    m  +  n(n2+l-m2)* 
li* +  1 

From  this  it  is  seen  that  no  one  of  all  the  possible  values 
of  sin  c,  in  either  (1)  or  (2),  can  be  numerically  greater  than 
unity. 

In  (1),  there  is  one  solution  only,  for  one  value  of  sin  c 
is  negative,  and  B  must  be  of  the  same  quadrant  with  b. 

In  (2),  there  are  two  solutions,  real  and  distinct,  in  sin  c, 


SPHERICAL   TRIANGLES.  103 

unless  n  (n2  +  1  —  m2)*  =  0,  i.e.,  unless  b  =  — ,  or  else  cos2a 

—  cos2b  =  sin  b  cos  A,  when  the  two  solutions  in  sin  c  analyti- 
cally coincide.  For,  with  this  possible  exception,  we  have 

proved  that  1  >    .      1  >  0,   and  sin  GI  4=  sin  c2 .     If  n  =  0, 

smc2 

I.e.,  if  b  =  — ,  the  two  analytically  coincident  solutions  in  sin  c 

give  two  real  triangles,  in  which  the  values  of  c  are  supple- 
mentary, as  shown  in  §  89. 

If  n2-fl  —  m2  —  0,  i.e.,  if  cos2a  —  cos2b  =  sin  b  cos  A,  the 
condition  sin  B<1,  is  violated,  and  the  solutions  are  not 
real.* 

It  might  seem  as  if  sin  Ci  ,  and  sin  c2 ,  would  give  two  values 
of  G!  and  of  c2,  which  taken  with  the  two  values  of  B,  obtained 
from  sin  B,  would  give  eight  triangles,  but  the  formulas  of 
§  96,  which  we  use  in  finding  C  and  c,  show  that  only  one 
value  of  c  can  consist  with  one  value  of  B.  (See  examples 
XI.  24,  25,  26.) 

We  may  now  consider  that  we  have 

b 


bc2A 
given 

B2C2 
required 

where  TT  >     1 

C2 

A 

We  may  find  B  by  the  formula, 

sinB  =  siDAsinb.  §79 

sin  a 

If  two  values  of  B  are  admissible,  they  are  supplements  of 
each  other  and  may  be  denoted  by  B!  and  B2.  When  B  is 
found,  c  and  C  may  be  obtained  from  the  formulas  : 

*  It  is  easily  shown  in  the  Theory  of  Functions  that  sin—1  x,  if  x  >  1, 
is  a  complex  angle. 


104  SPHERICAL   TRIGONOMETRY. 

A+B 


a  +  b 
C      Sm  —         A-B 


sn 


If  there  are  two  solutions  we  have  : 

.     A  +  B, 
Cl      Sm—          a-b 


sin 


.    A+B2 
Sm-2-         a-b 


with  two  similar  formulas  for  Ci  and  C2 . 
Case  6.     Given  A,  B,  and  a. 

(1)  If  the  inequalities,  7r>A>0,  7r>B>0,  7r>a>0, 
are  satisfied,  there  will  always  be  one  solution,  when  B  differs 

more  than  A  from  —  • 

LJ 

(2)  Unless  the  inequalities, 

sin  a  sin  B  _       .     .  _.  ^  .        ^  A  ^  TT 

: — —  -  [=  sm  b]  <  1,  TT  >     >  -  > 

sm  A       L  a       2 

or  else         —  >     >  0, 
and  TT  >  B  >  0, 

are  satisfied  when  A  differs  more  than  B  from  —  >  there  will 

be  no  solution  ;  and  when  they  are  satisfied,  and  A  and  B  thus 
related,  there  will  always,  necessarily,  be  two  solutions. 


SPHERICAL   TRIANGLES.  105 

These    statements    might   be    proved   similarly  as  in   the 
preceding  case,  taking  the  equation, 

.    ^      cos  A  +  cos  B  cos  C 
sin  C  =  -  .     p  -  •  §  81 

sin  D  cos  a 

But  it  is  not  necessary  to  do  this.  If  A,  B.  and  a  are  given, 
a',  b',  and  A'  of  the  polar  triangle  are  given.  If  the  polar 
triangle  has  two  solutions,  the  given  triangle  will  have  two 
solutions,  and  conversely.  The  polar  triangle  will  have  two 

solutions,  when  a  '  differs  more  than  b'  from  —  •  The  given 
triangle  will  therefore  have  two  solutions,  when  IT  —  A  differs 
more  than  TT  —  B  from  —•>  i.e.,  when  A  differs  more  than  B 

from  —  •     We  may  find  b  by  the  formula 

L 

sin  a  sin  B 

sin  b  =  --  :  —  T—  §  79 

sm  A 

and  if  two  values  of  b  are  admissible  they  are  supplements  of 
each  other,  and  may  be  denoted  by  bt  and  b2.  c  and  C  may 
be  found  from  the  formulas  : 

A  +  B 
c       Sm—         a-b 


Bn—  |- 

a  +  b  §96 

Sm~         A-B 


Similarly  if  there  are  two  solutions. 


106  SPHERICAL   TRIGONOMETRY. 


J>.     Examples.     XL 

Apply  the  conditions  of  possibility,  and  solve  the  following  triangles, 
having  given : 

1.  a  =  124°  12'. 5,  b  =  54°  18',  c  -  97°  12/.5. 

2.  a  =  74°  26',  b  =  85°  12',  c  =  29°  47'. 

3.  A  =  20°  10',   B  =  55°  52',  C  =  114°  20'. 

4.  A  =  130°,   B  =  110°,  C  =  80°. 

6.  a  =  35°  37',  b  =  59°  12',  C  =  124°  18'. 

6.  A  =  54°  55',  b  =  69°  25',  c  =  109°  46'. 

7.  A  =  26°  59',   B  =  39°  45',  c  -  154°  47'. 

8.  A  =  57°  35',   B  =  120°  48',  c  =  124°  18'. 

9.  a  =  50°,  b  =  40°,  A  =  80°. 

10.    a  =  150°  57',  b  =  134°  16',  A  =  144°  23'. 
•11.    A  =  113°  39',   B  =  123°  40',  a  =  65°  40'. 

12.  A  =  52°  50',   B  =  66°  07',  a  =  59°  28'. 

13.  Are  the  following  triangles  possible  ?     If  impossible,  what  condi- 
tions are  violated  ? 

(1)  a  =  160°  25',  b  =  92°  23',  c  =  64°  49'. 

(2)  A  -  137°  18',  B  =  123°  15',  C  =  74°  36'. 

(3)  a  =  2°  29',  b  =  37'.  5,  C  =  179°  48'. 

(4)  A  =  5°  41',   B  =  15°  50',  c  =  163°  34'. 

(5)  A  =  88°  49',  a  =  16°  34',   b  =  16°  30'. 

(6)  a  =  30°  08',   b  =  54°  03',  A  -  60°  01'. 

(7)  a  =  117°  29',  A  =  174°  17',   B  =  173°  25'. 

14.  Prove  that  the  polar  radii  of  the  inscribed  circle  of  a  triangle,  and 
of  the  circumscribed  circle  of  its  polar  triangle  are  complements  of  each 
other. 

15.  If  R,  Ra,  Rb,  Re,  are  the  polar  radii  of  the  circles  circumscribed 
about  the  triangles  ABC,"  BCA',  CAB',  ABC',  respectively,  where  A  A',  BB', 
CC',  are  diameters  of  the  sphere,  prove  that 

—  ctn  R  cos  S  =  ctn  Ra  cos  (S  —  A) 
=  ctnRbCos(S  —  B) 
=  ctnRccos(S  —  C) 


=  v—  cos  S  cos  (S  —  A)  cos  (S  —  B)  cos  (S  —  C). 
[=  L,  for  brevity.] 

16.  If  r,  ra,  rb,   rc,  are,  respectively,  the  polar  radii  of  the  circles 
inscribed  in  the  triangles  of  the  preceding  problem,  prove  that 


EXAMPLES.     XL  107 

tan  r  sin  s  =  tan  ra  sin  (s  —  a)  =  tan  rb  sin  (s  —  b)  =  tan  rc  sin  (s  —  c) 
=  Vsin  s  sin  (s  —  a)  sin  (s  —  b)  sin  (s  —  c). 
[=  I,  for  brevity.] 

Using  the  notation  of  examples  15  and  16,  prove  the  following  : 

17.  tan  r  tan  ra  tan  rb  tan  rc  =  1  2. 

18.  ctn  R  ctn  Ra  ctn  Rb  ctn  Rc  =  L2. 

19.  tan  ra  tan  rb  tan  rc  =  I  sin  s. 

20.  ctn  Ra  ctn  Rb  ctn  Rc  =  —  L  cos  S. 

4  L  sin  S 

21.  tan  ra  +  tan  rb  +  tan  rc  -  tan  r  -  — 


22.  tan  Ra  +  tan  Rb  +  tan  Rc  —  tan  R  =  2  ctn  r. 

23.  ctn  R  —  ctn  Ra  —  ctn  Rb  —  ctn  Rc  =  -  j  -- 

24.  Show  that  if  the  inequalities  of  §  101.  6.  Case  5.  (1),  are  satisfied, 

sin  A  sin  b  r       ,    _, 

-  [=  sin  B]  <  1. 
sin  a 

Show  that  under  §  101.  6.  Case  5.  (2),  the  two  analytically  coincident 
solutions  in  sin  c  : 

25.  Give  two  triangles  whose  sum  is  equal  to  a  lune,  if  n  =  0. 

26.  Violate  the  condition 

sin  A  sin  b  r 


for  real  triangles,  if  n2  +  1  —  m2  =  0. 

27.  Find  the  volume  V  of  an  oblique  parallelepiped,  having  given  the 
three  conterminous  edges  I,  m,  n,  and  the  three  angles  cr,  0,  7,  which 
the  edges  make  with  each  other. 

•In  the  following  problems  the  earth  is  assumed  to  be  a  sphere  whose 
radius  is  3963  miles. 

28.  Given  the  latitudes  and  longitudes  of  three  places  on  the  earth's 
surface,  show  how  to  find  the  area  of  the  triangle  of  which  the  given 
places  are  the  vertices. 

29.  Colorado  extends  from  37°  to  40°  N.,  and  from  102°  to  107°  W. 
What  is  its  area  in  statute  square  miles  ? 

30.  If  Pennsylvania  is  assumed  to  extend  from  39°  43'  to  42°  N.  ,  and 
from  75°  to  80°  35'  W.,  what  is  its  area  in.  statute  square  miles  ? 

31.  If  Ohio  is  assumed  to  extend  from  39°  to  41°  30'  N.,  and  from 
80°  35'  to  84°  40'  W.,  what  is  its  area  in  statute  square  miles  ? 

32.  Given  the  latitudes  and  longitudes  of  two  places  on  the  earth's 
surface,  show  how  to  find  the  shortest  distance  between  them. 


108  SPHERICAL   TRIGONOMETRY. 

33.  What  is  the  shortest  distance  in  statute  miles,  from  Cambridge, 
Mass.  (lat.  42°  23'  N.,  long.  77°  08'  W.),  to  Oberlin  (lat.  41°  17'  N.,  long. 
82°  11'  W.)? 

34.  What  is  the  shortest  distance  in  statute  miles  from  Philadelphia 
(lat.  39°  57'  N.,  long.  75°  10'  W.)  to  Oberlin  ? 

35.  Given  the  shortest  distance  between  two  places,  and  their  lati- 
tudes, show  how  to  find  the  difference  in  their  local  time. 

36.  The  shortest  distance  between  Sandy  Hook  (lat.  40°  28'  N.)  and 
Cork  Harbor  (lat.  51°  47'  N.)  is  2726  geographical  miles.     When  it  is 
6  A.M.  at  Sandy  Hook,  what  time  is  it  at  Cork  Harbor  ? 

37.  Given  0,  5,  and  t.      What  is  the  formula  for  the  position  angle  of 
the  astronomical  triangle  ? 

38.  At  Oberlin  Observatory,  the  declination  of  a  star  is  54°  25',  its 
hour  angle  10  h.  20  m.     What  is  its  position  angle  ? 

39.  Find  the  distance  between  the  sun  and  moon  at  Greenwich  mean 
noon  on  June  20,  1896,  when  their  respective  right  ascensions  are  5  h. 
58  m.  11.09s.  and  13  h.  58  m.  21.65  s.,  and  their  declinations  23°  27'  14" 
and  17°  06'  15".  3. 

40.  Find  the  distance  between  Uranus  and  Neptune  at  Greenwich 
mean  noon  on  March  31,  1896,  when  their  respective  right  ascensions  are 
15  h.  26  m.  59.06  s.  and  4  h.  58  m.  8s.,  and  their  declinations  18°  30'  34" 
and  21°  16'  40". 4. 

41.  Find  the  distance  between  Venus  and  Jupiter  at  Greenwich  mean 
noon  on  Sept.  10,  1896,  when  their  respective  right  ascensions  are  12  h. 
21  m.  38.16  s.  and  9  h.  55  m.  29.76  s.,  and  their  declinations  1°  09'  19". 9 
and  13°  27'  06".  3. 

42.  Find  the  moon's  right  ascension  and  declination  at  Greenwich 
mean  midnight  Oct.  29,  1896,  when  her  true  longitude  is  124°  53'  30". 7, 
and  the  obliquity  of  the  ecliptic  is  23°  27'  16".  78. 

43.  Find  the  declination  of  a  star  whose  longitude  is  47°  25'  and 
latitude  51°  36'.     (Take  e  =  23°  27'.) 

44.  Find  the  latitude  of    a   star  whose  longitude  is  27°  19',  and 
declination  32°  48'. 


ADVERTISEMENTS 


MATHEMATICS.  93 

The  Method  of  Least  Squares. 

With  Numerical  Examples  of  its  Application.  By  GEORGE  C.  COM- 
STOCK,  Professor  of  Astronomy  in  the  University  of  Wisconsin,  and 
Director  of  the  Washburri  Observatory.  8vo.  Cloth,  viii  +  68  pages. 
Mailing  price,  $1.05 ;  for  introduction,  $1.00. 

College  Requirements  in  Algebra:  A  Final  Review. 

By  GEORGE  PARSONS  TIBBETS,  A.M.,  Instructor  in  Mathematics, 
Williston  Seminary.  12mo.  Cloth.  46  pages.  By  mail,  55  cents ;  to 
teachers  and  for  introduction,  50  cents. 

Peirce's  Elements  of  Logarithms. 

With  an  explanation  of  the  author's  Three  and  Four  Place  Tables.  By 
Professor  JAMES  MILLS  PEIRCE,  of  Harvard  University.  12mo.  Cloth. 
80  pages.  Mailing  price,  55  cents :  for  introduction,  50  cents. 

Mathematical  Tables  Chiefly  to  Four  Figures. 

With  full  explanations.  By  Professor  JAMES  MILLS  PEIRCE,  of  Har- 
vard University.  12ino.  Cloth.  Mailing  price,  45  cents;  for  intro- 
duction, 40  cents. 

Elements  of  the  Differential  Calculus. 

With  numerous  Examples  and  Applications.  Designed  for  Use  as  a 
College  Text-Book.  By  W.  E.  BYERLY,  Professor  of  Mathematics, 
Harvard  University.  8vo.  273  pages.  Mailing  price,  $2.15;  for  intro- 
duction, $2.00. 

rpHE  peculiarities  of  this  treatise  are  the  rigorous  use  of  the 
Doctrine  of  Limits,  as  a  foundation  of  the  subject,  and  as 
preliminary  to  the  adoption  of  the  more  direct  and  practically 
convenient  infinitesimal  notation  and  nomenclature  ;  the  early 
introduction  of  a  few  simple  formulas  and  methods  for  integrat- 
ing ;  a  rather  elaborate  treatment  of  the  use  of  infinitesimals  in 
pure  geometry  ;  and  the  attempt  to  excite  and  keep  up  the  interest 
of  the  student  by  bringing  in  throughout  the  whole  book,  and  not 
merely  at  the  end,  numerous  applications  to  practical  problems  in 
geometry  and  mechanics. 

E.  H.  Moore,  Professor  of  Mathe- 1  ablest  text-book  on  the  calculus  yet 
matics,  University  of  Chicago :  The  I  written  by  an  American. 


94  MATHEMATICS. 

Elements  of  the  Integral  Calculus. 

Second  Edition,  revised  and  enlarged.  By  W.  E.  BYERLY,  Professor 
of  Mathematics  in  Harvard  University.  8vo.  xvi  +  383  pages.  Mail- 
ing price,  $2.15;  for  introduction,  $2.00. 

work  contains,  in  addition  to  the  subjects  usually  treated 
in  a  text-book  on  the  Integral  Calculus,  an  introduction  to. 
Elliptic  Integrals  and  Elliptic  Functions  ;  the  Elements  of  the 
Theory  of  Functions  ;  a  Key  to  the  Solution  of  Differential  Equa- 
tions ;  and  a  Table  of  Integrals. 


John  E.  Clark,  Prof,  of  Mathe- 
matics, Sheffield  Scientific  School  of 
Yale  University:  The  additions  to 
the  present  edition  seem  to  me  most 
judicious  and  to  greatly  enhance  its 


value  for  the  purposes  of  university 
instruction,  for  which  in  several  im- 
portant respects  it  seems  to  me  hetter 
adapted  than  any  other  American 
text-book  on  the  subject. 


An   Elementary   Treatise  on   Fourier's  Series, 

and  Spherical,  Cylindrical,  and  Ellipsoidal  Harmonics,  with  Appli- 
cations to  Problems  in  Mathematical  Physics. 

By  WILLIAM  E.  BYERLY,  Ph.D.,  Professor  of  Mathematics,  Harvard 
University.  8vo.  Cloth,  x  +  288  pages.  Mailing  price,  $3.15 ;  for 
introduction,  $3.00. 

HHHIS  book  is  intended  as  an  introduction  to  the  treatment  of 
some  of  the  important  Linear  Partial  Differential  Equations 
which  lie  at  the  foundation  of  modern  theories  in  physics,  and 
deals  mainly  with  the  methods  of  building  up  solutions  of  a 
differential  equation  from  easily  obtained  particular  solutions,  in 
such  a  manner  as  to  satisfy  given  initial  conditions. 


John  Perry,  Technical  College, 
Finfsbury,  London,  England:  Byer- 
ly's  book  is  one  of  the  most  useful 
books  in  existence.  I  have  read  it 
with  great  delight  and  I  am  happy 


to  say  that  although  it  seemed  to  be 
written  expressly  for  me,  one  of  my 
friends  who  is  a  great  mathemati- 
cian, seems  as  delighted  with  it  as  I 
am  myself. 


A    Short    Table   Of  Integrals.       Revised  and  Enlarged. 

By  B.  O.  PEIRCE,  Prof.  Math.,  Harvard  Univ.  32  pages.  Mailing 
price,  15  cents.  Bound  also  with  Byerly's  Calculus. 

B  LI  er Iyfs  Syllabi. 

Each,  8  or  12  pages,  10  cents.  The  series  includes,— Plane  Trigonom- 
etry, Plane  Analytical  Geometry,  Plane  Analytic  Geometry  (Advanced 
Course),  Analytical  Geometry  of  Three  Dimensions,  Modern  Methods 
in  Analytic  Geometry,  the  Theory  of  Equations. 


MATHEMATICS.  95 

Directional  Calculus. 

By  E.  W.  HYDE,  Professor  of  Mathematics  in  the  University  of  Cincin- 
nati. 8vo.  Cloth.  xii  + 247  pages,  with  blank  leaves  for  notes.  Price 
by  mail,  $2.15;  for  introduction,  $2.00. 

HHHIS  work  follows,  in  the  main,  the  methods  of  Grassmann's 
Ausdehnungslehre,  but  deals  only  with  space  of  two  and  three 
dimensions.  The  first  two  chapters  which  give  the  theory  and 
fundamental  ideas  and  processes  of  his  method,  will  enable  students 
to  master  the  remaining  chapters,  containing  applications  to  Plane 
and  Solid  Geometry  and  Mechanics ;  or  to  read  Grassmann's  original 
works.  A  very  elementary  knowledge  of  Trigonometry,  the  Differ- 
ential Calculus  and  Determinants,  will  be  sufficient  as  a  preparation 
for  reading  this  book. 


Daniel  Carhart,  Prof,  of  Mathe- 
matics, Western  University  of  Penn- 
sylvania :  I  am  pleased  to  note  the 
success  which  has  attended  Professor 


Hyde's  efforts  to  bring  into  more 
popular  form  a  branch  of  mathemat- 
ics which  is  at  once  so  abbreviated  in 
form  and  so  comprehensive  in  results. 


Elements  of  the  Differential  and  Integral  Calculus. 

With  Examples  and  Applications.  By  J.  M.  TAYLOR,  Professor  of 
Mathematics  in  Colgate  University.  8vo.  Cloth.  249  pages.  Mailing 
price,  $1.95;  for  introduction,  $1.80. 

rpHE  aim  of  this  treatise  is  to  present  simply  and  concisely  the 
fundamental  problems  of  the  Calculus,  their  solution,  and  more 
common  applications. 

Many  theorems  are  proved  both  by  the  method  of  rates  and  that 
of  limits,  and  thus  each  is  made  to  throw  light  upon  the  other. 
The  chapter  on  differentiation  is  followed  by  one  on  direct  integra- 
tion and  its  more  important  applications.  Throughout  the  work 
there  are  numerous  practical  problems  in  Geometry  and  Mechanics, 
which  serve  to  exhibit  the  power  and  use  of  the  science,  and  to 
excite  and  keep  alive  the  interest  of  the  student.  In  February,  1891, 
Taylor's  Calculus  was  found  to  be  in  use  in  about  sixty  colleges. 


The  Nation,  New  York :  In  the 
first  place,  it  is  evidently  a  most 
carefully  written  book. . . .  We  are 
acquainted  with  no  text-book  of  the 
Calculus  which  compresses  so  much 
matter  into  so  few  pages,  and  at  the 
same  time  leaves  the  impression  that 


all  that  is  necessary  has  been  said. 
In  the  second  place,  the  number  of 
carefully  selected  examples,  both  of 
those  worked  out  in  full  in  illustra- 
tion of  the  text,  and  of  those  left  for 
the  student  to  work  out  for  himself, 
is  extraordinary. 


MATHEMATICS. 


Elements  of  Solid  Geometry. 

By  ARTHUR  LATHAM  BAKER,  Professor  of  Mathematics,  University  of 
Rochester.  12mo.  Cloth,  xii  + 126  pages.  Mailing  price,  90  cents; 
for  introduction,  80  cents. 

distinctive  features  of  this  work  are  improved  notation, 
tending  to  simplify  the  text  and  figures  ;  improved  diagrams, 
particular  attention  being  paid  to  the  perspective  of  the  figures  ; 
clear  presentation,  each  part  of  the  discussion  being  presented 
under  a  distinct  heading  ;  generalized  conceptions,  which  is  the 
principal  feature  of  the  work,  the  general  theorems  for  the  frustum 
of  a  pyramid  being  first  worked  out,  and  then  the  pyramid,  cone, 
prism,  and  cylinder  discussed  as  special  cases  of  the  pyramidal 
frustum  and  of  the  prismatoid.  The  essential  unity  of  the  sub- 
ject is  constantly  impressed  upon  the  reader. 


Benjamin  G.  Brown,  Professor  of 
Mathematics  in  Tufts  College :  It  is 
a  most  excellent  book.  I  have  never 


used  a  book  for  the  first  time  with 
greater  satisfaction. 


Elementary  Co-ordinate  Geometry. 


By  W.  B.  SMITH,  Professor  of  Mathematics,  Missouri  State  University. 
8vo.    Cloth.    312  pages.    Mailing  price,  $2.15;  for  introduction,  $2.00. 


book  is  spoken  of  as  the  most  exhaustive  work  on  the 
subject  yet  issued  in  America  ;  and  in  colleges  where  an  easier 
text-book  is  required  for  the  regular  course,  this  will  be  found  of 
great  value  for  post-graduate  study. 


Wm.  G.  Peck,  late  Prof,  of  Math- 
ematics and  Astronomy,  Columbia 
College:  Its  well  compacted  pages 


contain  an  immense  amount  of  mat' 
ter,  most  admirably  arranged.  It  is 
an  excellent  book. 


Theory  of  the  Newtonian  Potential  Functions. 

By  B.  O.  PEIRCE,  Professor  of  Mathematics  and  Physics,  in  Harvard 
Univ.    8vo.    Cloth.    154  pages.    Mailing  price,  $1.60;  for  introd.  $1.50. 


book  gives  as  briefly  as  is  consistent  with  clearness  so 
much  of  that  theory  as  is  needed  before  the  study  of  standard 
works  on  Physics  can  be  taken  up  with  advantage.     A  brief  treat- 
ment of  Electrokinematics  and  many  problems  are  included. 


MATHEMATICS. 


97 


Academic  Trigonometry:   piane  and  spherical. 


By  T.  M.  BLAKSLEE, 

Moines  College,  Iowa.     12mo.     Cloth. 


Ph.D.  (Yale),  Professor  of  Mathematics  in  Des 
.     12mo.     Cloth.     33  pages.     Mailing  price,  30 
cents;  for  introduction,  25  cents. 

fPHE  Plane  and  Spherical  portions  are  arranged  on  opposite  pages. 

The  memory  is  aided  by  analogies,  and  it  is  believed  that  the 

entire  subject  can  be  mastered  in  less  time  than  is  usually  given  to 

Plane  Trigonometry  alone,  as  the  work  contains  but  29  pages  of  text 

The  Plane  portion  is  compact,  and  complete  in  itself. 

Examples  of  Differential  Equations. 

By  GEORGE  A.  OSBORNE,  Professor  of  Mathematics  in  the  Massachu- 
setts Institute  of  Technology,  Boston.  12mo.  Cloth,  vii  +  50  pages. 
Mailing  Price,  60  cents;  for  introduction,  50  cents. 

A  SERIES  of  nearly  three  hundred  examples  with  answers,  sys- 
"^  tematically  arranged  and  grouped  under  the  different  cases, 
and  accompanied  by  concise  rules  for  the  solution  of  each  case. 

Selden  J.  Coffin,  Prof,  of  Astron-  1  ance  is  most  timely,  and  it  supplies 
omy,  Lafayette  College  :  Its  appear-  |  a  manifest  want. 

Determinants. 

The  Theory  of  Determinants:  an  Elementary  Treatise.  By  PAUL  H. 
HANUS,  B.S.,  recently  Professor  of  Mathematics  in  the  University  of 
Colorado,  now  Assistant  Professor,  Harvard  University.  8vo.  Cloth. 
viii  +  217  pages.  Mailing  price,  $1.90  ;  for  introduction,  $1.80. 


book  is  written  especially  for  those  who  have  had  no  pre- 
vious  knowledge  of  the  subject,  and  is  therefore  adapted  to 
self-instruction  as  well  as  to  the  needs  of  the  class-room.  The 
subject  is  at  first  presented  in  a  very  simple  manner.  As  the 
reader  advances,  less  and  less  attention  is  given  to  details. 
Throughout  the  entire  work  it  is  the  constant  aim  to  arouse 
and  enliven  the  reader's  interest,  by  first  showing  how  the  various 
concepts  have  arisen  naturally,  and  by  giving  such  applications  as 
can  be  presented  without  exceeding  the  limits  of  the  treatise. 


William  G.  Peck,  late  Prof,  of 
Mathematics,  Columbia  College, 
N.  Y. :  A  hasty  glance  convinces  me 
that  it  is  an  improvement  on  Muir. 


T.  W.  Wright,  Prof,  of  Mathemat- 
ics, Union  Univ.,  Schenectady,  N.Y.: 
It  fills  admirably  a  vacancy  in  our 
mathematical  literature,  and  is  a 
very  welcome  addition  indeed. 


98  MATHEMATICS. 

Analytic  Geometry. 

By  A.  S.  HARDY,  Ph.D.,  recently  Professor  of  Mathematics  in  Dart- 
mouth College,  and  author  of  Elements  of  Quaternions.  8vo.  Cloth. 
xiv  +  239  pages.  Mailing  price,  $1.60;  for  introduction,  $1.50. 


work  is  designed  for  the  student,  not  for  the  teacher.     It 
is  hoped  that  it  will  prove  to  be  a  text-book  which  the  teacher 
will  wish  to  use  in  his  class-room,  rather  than  a  book  of  reference 
to  be  placed  on  his  study  shelf. 


Oren  Root,  Professor  of  Mathe- 
matics, Hamilton  College :  It  meets 
quite  fully  my  notion  of  a  text  for 
our  classes. 

John  E.  Clark,  Professor  of  Mathe- 


matics, Sheffield  Scientific  School  of 
Yale  College :  I  need  not  hesitate  to 
say,  after  even  a  cursory  examina- 
tion, that  it  seems  to  me  a  very  at- 
tractive book. 


Elements  of  Quaternions. 

By  A.  S.  HARDY,  Ph.D.,  recently  Professor  of  Mathematics  in  Dart- 
mouth College.  Second  edition  revised.  Crown.  8vo.  Cloth,  viii 
+  234  pages.  Mailing  price,  $2.15 ;  for  introduction,  $2.00. 

HPHE  chief  aim  has  been  to  meet  the  wants  of  beginners  in  the 
class-room. 

Elements  of  the  Calculus. 

By  A.  S.  HARDY,  Ph.D.,  recently  Professor  of  Mathematics  in  Dart- 
mouth College.  8vo.  Cloth.  xi  + 239  pages.  Mailing  price,  $1.60; 
for  introduction,  $1.50. 

Part  I.,  Differential  Calculus,  occupies  166  pages.    Part  II.,  Integral 
Calculus,  73  pages. 

HPHIS  text-book  is  based  upon  the  method  of  rates.  From  the 
author's  experience  in  presenting  the  Calculus  to  beginners, 
the  method  of  rates  gives  the  student  a  more  intelligent,  that  is,  a 
less  mechanical,  grasp  of  the  problems  within  its  scope  than  any 
other.  No  comparison  has  been  made  between  this  method  and 
those  of  limits  and  of  infinitesimals.  This  larger  view  of  the 
Calculus  is  for  special  or  advanced  students,  for  whom  this  work 
is  not  intended. 


Ellen  Hayes,  Professor  of  Mathe- 
matics, Wellesley  College :  I  have 
found  it  a  pleasure  to  examine  the 
book.  It  must  commend  itself  in 
many  respects  to  teachers  of  Cal- 
culus. 


J.  B.  Coit,  Professor  of  Mathe- 
matics, Boston  University :  The 
treatment  of  the  first  principles  of 
Calculus  by  the  method  of  rates  is 
eminently  clear. 


14  DAY  USE 

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THE  UNIVERSITY  OF  CALIFORNIA  ^IBRARY 


